Introduction to the representation theory of algebrasmath0.bnu.edu.cn/~liuym/Book-representation_theory.pdfbe Introduction to the theory of algebras, which are nite dimensional over

Introduction to the representation theory of

algebras

Michael Barot

January 25, 2012

ii

iii

To Angelica

Preface

The aim of these notes is to give a brief and elementary introduction tothe representation theory of finite-dimensional algebras. The notes origi-nated from an undergraduate course I gave in two occasions at UniversidadNacional Autonoma de Mexico.

The plan of the course was to try to cope with two competing demands: toexpect as little as possible and to reach as much as possible: to expect onlylinear algebra as background and yet to make way to substantial and centralideas and results during its progress.

Therefore some crucial decisions were necessary. We opted for the modelcase rather than the most general situation, for the most illustrating examplerather than the most extravagant one. We sought a guideline through thisvast field which conducts to as many important notions, techniques andquestions as possible in the limited space of a one-semester course. So, itis a book written from a specific point of view and the title should reallybe Introduction to the theory of algebras, which are finite dimensional oversome algebraically closed field.

The book starts with the most difficult chapter: matrix problems. Concep-tually there is little to understand in that chapter but it requires a consid-erable effort from the reader to follow the argumentation within. However,this chapter is central: it prepares all the main examples which later willguide through the rest. In the following two chapters we consider the mainlanguages of representation theory. Since there are several competing lan-guages in representation theory, a considerable amount of our effort is di-rected towards mastering and combining all of them. As you will see each ofthese languages has its own advantages and it therefore enables the readernot only to consult the majority of all research articles in the field, it alsoenriches the way we may thick about the notions themselves.

The rest of the book is devoted to gain structural insight into the categories

vi

of modules of a given algebra. In the chapter about module categories someolder results are proved, whereas in the next four chapters more recent devel-opments are discussed. The last chapter is more of an open minded collec-tion concerning indecomposable modules, the building bricks of the modulecategories, with respect to certain invariants, called dimension vectors.

My thanks go to Jan Schroer, for it was him who suggested to write jointlya book on representation theory but unfortunately gave up on it, to JurajHromkovic who invited me to ETH Zurich for a sabbatical in 2010 and madethus possible to finish the manuscript, to Manuela Tschabold who tirelesslyread the whole manuscript carefully eliminating thus many errors and mis-print, to Karin Baur who took it as a base for a course in 2011 suggestingmany improvements and to Angelica Herrera Loyo for her patience.

I apologize to the reader for all the errors which still remain and for thepoor english in which it is written.

Michael Barot

Instituto de MatematicasUniversidad Nacional Autonoma de Mexico

January 2012

Contents

1 Matrix problems 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 The two subspace problem . . . . . . . . . . . . . . . . . . . . 3

1.3 Decomposition into indecomposables . . . . . . . . . . . . . . 5

1.4 The Kronecker problem . . . . . . . . . . . . . . . . . . . . . 7

1.5 The three Kronecker problem . . . . . . . . . . . . . . . . . . 12

1.6 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Representations of quivers 17

2.1 Quivers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.3 Categories and functors . . . . . . . . . . . . . . . . . . . . . 22

2.4 The path category . . . . . . . . . . . . . . . . . . . . . . . . 26

2.5 Equivalence of categories . . . . . . . . . . . . . . . . . . . . . 29

2.6 A new example . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Algebras 37

3.1 Definition and generalities about algebras . . . . . . . . . . . 37

3.2 The path algebra . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.3 Quotients by ideals . . . . . . . . . . . . . . . . . . . . . . . . 42

3.4 Idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.5 Morita equivalence . . . . . . . . . . . . . . . . . . . . . . . . 47

3.6 The radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

vii

viii CONTENTS

3.7 The quiver of an algebra . . . . . . . . . . . . . . . . . . . . . 53

3.8 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4 Module categories 61

4.1 The categorical point of view . . . . . . . . . . . . . . . . . . 61

4.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.3 Kernels and cokernels . . . . . . . . . . . . . . . . . . . . . . 64

4.4 Unique decomposition . . . . . . . . . . . . . . . . . . . . . . 69

4.5 Projectives and injectives . . . . . . . . . . . . . . . . . . . . 72

4.6 Projective covers and injective hulls . . . . . . . . . . . . . . 78

4.7 Simple modules . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5 Elements of homological algebra 85

5.1 Short exact sequences . . . . . . . . . . . . . . . . . . . . . . 85

5.2 The Baer sum . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.3 Resolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.4 Long exact sequences . . . . . . . . . . . . . . . . . . . . . . . 101

6 The Auslander-Reiten theory 107

6.1 The Auslander-Reiten quiver and the infinite radical . . . . . 107

6.2 Example: the Kronecker algebra . . . . . . . . . . . . . . . . 109

6.3 The Auslander-Reiten translate . . . . . . . . . . . . . . . . . 117

6.4 Auslander-Reiten sequences . . . . . . . . . . . . . . . . . . . 121

7 Knitting 129

7.1 Source and sink maps . . . . . . . . . . . . . . . . . . . . . . 129

7.2 The knitting technique . . . . . . . . . . . . . . . . . . . . . . 133

7.3 Knitting with dimension vectors . . . . . . . . . . . . . . . . 136

7.4 Limits of knitting . . . . . . . . . . . . . . . . . . . . . . . . . 139

7.5 Preprojective components . . . . . . . . . . . . . . . . . . . . 143

8 Combinatorial invariants 149

CONTENTS ix

8.1 Grothendieck group . . . . . . . . . . . . . . . . . . . . . . . 149

8.2 The homological bilinear form . . . . . . . . . . . . . . . . . . 152

8.3 The Coxeter transformation . . . . . . . . . . . . . . . . . . . 155

8.4 Quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . 158

8.5 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

9 Indecomposables and dimensions 169

9.1 The Brauer-Thrall conjectures . . . . . . . . . . . . . . . . . . 169

9.2 Gabriel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 172

9.3 Reflection functors . . . . . . . . . . . . . . . . . . . . . . . . 175

9.4 Kac’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 179

9.5 Tame and wild . . . . . . . . . . . . . . . . . . . . . . . . . . 181

9.6 Module varieties . . . . . . . . . . . . . . . . . . . . . . . . . 187

x CONTENTS

Chapter 1

Matrix problems

“Matrix problems” provide easy examples for the phenomena we are inter-ested in to study. At the same time they give us a powerful machinery toexplicitly calculate these examples and study several main aspects of im-portance in the theory of representations of algebras. They however do notprovide a nice and deep theory as we will see later.

We will therefore not pursue the theory of “matrix problems” formally butapproach it intuitively by looking at a series of examples. As you will see allthese examples have certain common features and all will be of importancethroughout the whole book.

1.1 Introduction

Roughly speaking, we look at a set M and an equivalence relation on Mand ask for a normal form, that is, an explicit list of representatives ofthe equivalence classes. The following example is not only illustrative butessential for all others.

Example 1.1. Let M be the set of all matrices with entries in a fixed fieldK and define M ∼ N if there exist invertible matrices (of suitable size) Uand V such that N = UMV −1.

We know from linear algebra that any matrix M can be transformed byelementary row and column transformations into[

1r 00 0

], (1.1)

1

1 Matrix problems

where 1r stands for an identity matrix of size r and 0 for the zero matrixof some size. This transformations correspond to the multiplication with Uand V respectively. Hence in each equivalence class there exists at least onematrix of the form (1.1). If we interpret M as the matrix corresponding to alinear map f between two fixed vector spaces, then the equivalence class ofM corresponds to all matrices which represent f in different basis of thesevector spaces. The number r (the size of the identity matrix in the upperleft corner) is just the rank of M . Since the rank and also the size of Mitself are invariant under change of basis, we see that only one matrix of theform (1.1) belongs to each equivalence class. The form (1.1) is determinedby three integers r and m,n, where m× n is the size of the matrix. ♦

Comments 1.2 (a) The set M of Example 1.1 is a disjoint union M =⋃(m,n)∈N2 Mm,n indexed over pairs of natural numbers, where Mm,n =

Km×n is the set of all matrices of size m × n. Each equivalence class iscontained in a specific set Mm,n.

We will always consider 0 to be an element of N. The matrix of size m× 0(resp. 0× n) are those matrices corresponding to the linear maps when oneof the two vector spaces is the zero space. Such matrices have no entries butthey do have a size. As we will see it will be very convenient to considersuch strange matrices.

(b) The equivalence classes can be viewed similarly as the orbits undera group action, with the important difference that the action is given bya disjoint union of groups G =

⋃m,n∈NGm,n, where Gm,n = GLm(K) ×

GLn(K). That is, we have various actions Gm,n ×Mm,n →Mm,n, given by(U, V ) ·M = UMV −1. The importance of looking at all sizes at the sametime will become clearer later.

In the following we will look at some concrete problems of gradually in-creasing difficulty. We will be able to solve the first two problems but failin the last one. The solutions encountered are of outstanding interest forrepresentation theory and that is the reason why we start with them.

Exercises

1.1.1 Let M be the set of all square matrices with entries in the algebraicallyclosed field K and define the equivalence relation by conjugacy, that is, M ∼ N

2

1.2 The two subspace problem

holds if and only if there exists an invertible matrix U such that N = UMU−1.What are then the normal forms known from linear algebra? How would you divideM into disjoint subsets? Describe the group actions in these subsets.

1.1.2 Let M be the set of all matrices with entries in the field K. Define theequivalence relation by one-sided transformation, that is, M ∼ N if and only ifthere exists an invertible matrix U such that N = UM . The normal form is knownas reduced row-echelon form in linear algebra. How many normal forms of size 3×3exist if K is the field with two elements?

1.2 The two subspace problem

The two subspace problem is given by the set M of all pairs of matriceswith the same number of rows under the equivalence relation: (A,B) ∼(A′, B′) if there exists invertible matrices U, S and T such that A′ = UAS−1

and B′ = UBT−1.

We write [A|B] instead of (A,B) for such a pair and think of it as one bigmatrix divided into two vertical stripes. The allowed transformations forthis problem consist of row and column transformations which do not changethe given equivalence class, but are powerful enough to ensure transitivitywithin - in our given problem arbitrary row transformations of [A|B] andarbitrary column transformations within each stripe separately.

Notice that the set M is naturally divided as M =⋃m,n′,n′′∈N Mm,n′,n′′

where Mm,n′,n′′ = Km×n′ ×Km×n′′ .

Example 1.3. Before we enter the general case, we consider a concreteexample. Let

M =1

4

7

2

5

8

3

6

9

.

We first give the sequence of transformations and comment afterwards.

R−41

4

7

2

5

8

3

6

9

∼U

−7

1

0

7

2

−3

8

3

−6

9

∼ (−3)·

1

0

0

2

−3

−6

3

−6

−12

∼R6

1

0

0

2

1

−6

3

2

−12

∼

−21

0

0

2

1

0

3

2

0

∼

(−1)·

1

0

0

0

1

0

−1

2

0

∼ R21

0

0

0

1

0

1

−2

0

∼

−2

1

2

0

0

1

0

1

0

0

∼1

0

0

0

1

0

1

0

0

3

1 Matrix problems

The arrows on the left of the matrix indicate the elementary row transforma-tion of adding a multiple of a row to another row, the factor is indicated as asmall number. Similarly an arrow above the matrix indicates an elementarycolumn transformation. There are only two more elementary transforma-tions, namely the multiplication of the second row with −3 in the third stepand the multiplication of the last column by −1 in the fifth step. The lastmatrix is the normal form we sought. ♦

Applying (1.1) to the block A in [A|B] (by this, we mean that we use rowand column transformation in the left block of the matrix [A|B]), we seethat [A|B] is equivalent to a matrix of the form[

1r 00 0

∣∣∣∣ B1

B2

].

We apply (1.1) again to B2 and see that [A|B] is equivalent to1r 00 00 0

∣∣∣∣∣∣B1,1 B1,2

1c 00 0

.By adding multiples of rows of the second row-stripe to the first row-stripewe certainly do not change anything in the first two column-stripes andachieve that B1,1 = 0. Finally, we apply (1.1) to B1,2. Observe that the rowtransformations will disturb the identity block 1r, but this can by correctedusing column transformations in the first column stripe. Hence we obtain anormal form

1a 0 00 1b 00 0 00 0 0

∣∣∣∣∣∣∣∣0 1a 00 0 01c 0 00 0 0

, (1.2)

where a + b = r. But the normal form depends on more than the threenon-negative parameters a, b c, namely also on the number of zero rows atthe bottom and the number of zero columns in the third and sixth verticalstripe. We hence have proved the following result.

Proposition 1.4. The two subspace problem has the normal form given in(1.2).

4

1.3 Decomposition into indecomposables

Exercises

1.2.1 Determine which elements of the two subspace problem in the following listare equivalent.

M1 =

[1 11 1

∣∣∣∣ 1 00 0

], M2 =

[1 11 0

∣∣∣∣ 0 01 0

], M3 =

[1 01 0

∣∣∣∣ 1 01 0

],

M4 =

[1 11 0

∣∣∣∣ 1 00 0

], M5 =

[5 11 1

∣∣∣∣ 1 30 0

], M6 =

[5 11 1

∣∣∣∣ 1 31 0

].

1.3 Decomposition into indecomposables

As the problems grow larger the complexity in the division of the matriceswill increase. We therefore seek a procedure to simplify the notation andenlarge our understanding.

Given two matrices A ∈ Km×n and C ∈ Kr×s we define a new matrixA⊕ C ∈ K(m+r)×(n+s) by

A⊕ C =

[A 00 C

].

If two elements (A,B) and (C,D) of the two subspace problem are given,we define a new one by:

(A,B)⊕ (C,D) = (A⊕ C,B ⊕D)

and call it the direct sum of (A,B) and (C,D).

We can view ⊕ as a binary operation

M ×M →M , (M,N) 7→M ⊕N,

which satisfies Mm,n′,n′′ ⊕Mp,q′,q′′ ⊆Mm+p,n′+q′,n′′+q′′ . We can rewrite thisas Md ⊕Me ⊆Md+e for each d, e ∈ N3. We shall thus write M0 instead ofM0,0,0.

We call a pair (A,B) indecomposable if it does not lie in the equivalenceclass of two strictly smaller pairs, that is, if (A,B) is equivalent to (A′, B′)⊕(A′′, B′′) then either (A′, B′) ∈M0 or (A′′, B′′) ∈M0 but not both.

Notice that the unique element of M0 is, by definition, not indecomposable.Notice also that indecomposbility is a property which is invariant in eachequivalence class.

We can now decompose (1.2) into indecomposables.

5

1 Matrix problems

Proposition 1.5. There are only 6 indecomposable normal forms for thetwo subspace problem, namely the following:

[|], , , [1|], [|1], [1|1], (1.3)

where [|], and stand for the unique element in M1,0,0, M0,1,0 andM0,0,1 respectively.

Proof. We start rearranging the right block of (1.2) to get an equivalentmatrix and then decompose as follows:

1a 0 00 1b 00 0 00 0 0

∣∣∣∣∣∣∣∣1a 0 00 0 00 1c 00 0 0

= [1a|1a]⊕

1b 00 00 0

∣∣∣∣∣∣0 01c 00 0

= [1a|1a]⊕ [1b|]⊕

[00

∣∣∣∣ 1c 00 0

]= [1a|1a]⊕ [1b|]⊕ [|1c]⊕ [0|0].

But [1a|1a] = [1|1]⊕ [1|1]⊕ . . .⊕ [1|1] with a summands on the right handside. We abbreviate this by [1|1]a. Similarly [1b|] = [1|]b and [|1c] = [|1]c.Finally the zero matrix [0|0] decomposes as

[0|0] = [|]r ⊕ s′ ⊕ s′′,

where r is the number of rows of [0|0] and s′ (resp. s′′) is the number ofcolumns of the left (resp. right) stripe. Hence we get altogether the followingdecomposition

M = [1|1]a ⊕ [1|]b ⊕ [|1]c ⊕ [|]r ⊕ s′ ⊕ s′′.

Exercises

1.3.1 Decompose each of the following matrices into indecomposables.

M1 =

[1 11 1

∣∣∣∣ 1 00 0

], M2 =

[1 10 0

∣∣∣∣ 1 10 0

], M3 =

[1 01 0

∣∣∣∣ 0 11 0

].

1.3.2 Calculate all indecomposables for the dual problem of the two subspaceproblem: consider pairs of matrices (A,B) with the same number of columns underthe equivalence relation: (A,B) ∼ (A′, B′) if there exist invertible U, V, T such thatA′ = UAT−1 and B′ = V BT−1.

6

1.4 The Kronecker problem

1.3.3 Consider the three subspace problem consisting of triples of matrices(A,B,C) with the same number of rows under the equivalence relation: (A,B,C) ∼(A′, B,C ′) if there exist invertible matrices R,S, T, U such that A′ = RAU−1,B′ = SBU−1 and C ′ = TCU−1. Define indecomposability and then calculate acomplete list of indecomposables (up to equivalence). You should get a list of 12indecomposables.


The Kronecker problem is the matrix problem of pairs of matrices (A,B)of the same size under the equivalence relation: (A,B) ∼ (A′, B′) if thereexist invertible matrices U, T such that A′ = UAT−1 and B′ = UBT−1.The direct sum is defined in analogy to the two subspace problem: (A,B)⊕(C,D) = (A⊕ C,B ⊕D) and induces the notion of indecomposablity.

This problem is substantially more complicated. We will have to manipulatecarefully until we get the final list of indecomposables, see Proposition 1.6.However, we shall not start from the solution but show how it can be found.

For this, we think the pairA,B again as one matrix with a vertical separationin the middle and write [A|B] instead of (A,B). But we should not forgetthat the columns of A are coupled with the columns of B. Given such a pair[A|B], we start again by reducing A to normal form (1.1) but introduce aslight modification by putting the identity matrix to the upper right corner.But then the matrix B should be divided into 4 blocks of the correspondingsize just as in the left vertical stripe:[

0 10 0

∣∣∣∣ C DE F

],

in particular D is a square matrix. If we just focus on the third and fourthvertical stripe, what are the allowed transformations? Well, we are allowedto do arbitrary row transformations separately in each horizontal stripe,but by doing so in the upper one, we spoil the identity matrix in the secondvertical stripe. This can be fixed by applying the inverse transformationto the second vertical stripe (and consequently to the forth, remember thecoupling). Therefore D is transformed by conjugation. And clearly we arealso allowed to apply arbitrary column transformations to the third verticalstripe (and the first, which is zero and thus doesn’t bother us). But we areallowed to do even more. We also can add multiples of the lower stripe tothe upper stripe and multiples of the left stripes (first and third) to the rightstripes (second and forth).

7

1 Matrix problems

This is a new challenging problem! We shall call it momentarily the cou-pled 4-block, just to have a name for it, since it will appear again. Considersuch a coupled 4-block and apply (1.1) to the lower left corner, i.e. to E,again slightly different. The division in the left vertical and lower horizontalstripe is then prolonged throughout the matrix and we obtain:C ′ C ′′ D′′

0 0 F ′

1 0 F ′′

.By adding multiples of the lower to the upper stripe we can force C ′ = 0and similarly by adding multiples from the left to the right, we can forceF ′′ = 0. Hence, we obtain[

C DE F

]∼

0 C ′′ D′

0 0 F ′

1 0 0

=

[C ′′ D′

0 F ′

]⊕[

1].

That is, we could get rid of the lower right block by splitting off a directsummand, composed itself of indecomposables

[1], the one standing in

the lower left corner. Now, apply (1.1) to F ′, but remember that columntransformations in the right stripe are coupled with row transformations inthe upper stripe, and therefore the vertical division coming from F ′ takesover to a horizontal division of the upper stripe:

[C ′′ D′

0 F ′

]∼

C1 D1 D2

C2 D3 D4

0 0 10 0 0

∼C1 D1 0

C2 D3 00 0 10 0 0

=

C1 D1 0

C2 D3 00 0 1

⊕ [ ] ,where

[ ], stands for a lower zero row with no columns. Observe that D1

is a square matrix again and that we can not split off a direct summandcontaining the 1 in the third vertical stripe because it is coupled with thesecond horizontal stripe. What are the allowed transformations in the fourblocks C1, C2, D1 and D3? To see this we recall how the four blocks areembedded in the original problem:0 1 0 C1 D1 0

0 0 1 C2 D3 00 0 0 0 0 1

. (1.4)

Recall that in the original 4-block the row transformations of the block Dwere coupled with the column transformations of the same block. Hence the

8


row transformations of D1 are also coupled with column transformations ofthe same block. Furthermore we can still do any column transformation inthe left stripe and add multiples of those columns to the right.

How about row transformations in the second horizontal stripe? They arecoupled with column transformations in the third vertical stripe by the entry1 in position (2, 3) in (1.4), see the next figure, where the numbers [1] and[2] refer to those two coupled transformations.

[1]6?[3]6?

[2]-

[2]-0 1 0 C1 D1 0

0 0 1 C2 D3 00 0 0 0 0 1

Column transformations in the third vertical stripe, [2], would spoil theachieved 1 in position (3, 6) – but we can repair that damage easily by rowtransformations in the third horizontal stripe [3]. This shows that, indeedwe can do arbitrary row transformations in the second horizontal stripe.

Similarly, if we want to add multiples from the second horizontal stripe to thefirst horizontal stripe, [1] in the next figure, then this action is coupled withthe the column transformation of adding multiples of the second verticalstripe to the third, [2].

[1]K[3]K

K

[2]j

[2]j0 1 0 C1 D1 0

0 0 1 C2 D3 00 0 0 0 0 1

Now transformation [2] would spoil the achieved blocks 0 in positions (1, 6)and (2, 6) but that damage can be repaired by adding multiples of the thirdhorizontal stripe to the upper ones, [3]. This shows that indeed we can addmultiples of the second to the first row.

So, after all we see that the four matrices C1 = C1, D1 = D1, E1 = C2 andF1 = D3 constitute again a coupled 4-block.

As it seems, we just can proceed iterating easily. But notice the shift inthe division: the original horizontal division between blocks has gone to thelower border, and therefore in the next step the indecomposables split offare of the form [

1 00 1

]and

[01

].

9

1 Matrix problems

In the subsequent steps the indecomposables which might be split off are:1 0 00 1 00 0 1

, 0 0

1 00 1

, . . . ,

1 0 · · · 00 1 · · · 0...

......

0 0 · · · 1

,

0 · · · 01 · · · 0...

...0 · · · 1

.The corresponding indecomposables of the Kronecker problem are of theform

[J(m, 0) | 1m], for m ≥ 0, (1.5)

where J(m, 0) is the matrix of size m ×m having as only non-zero entriesJ(m, 0)ab = 1 if a = b− 1 and[

1nz>

∣∣∣∣ z>1n], for n ≥ 1, (1.6)

where z denotes a zero column of appropriate size and M> the transpose ofa matrix M . In each step the remaining matrix is of the form

Ci Di 0 0 · · · 0Ei Fi 0 0 · · · 00 0 1 0 · · · 00 0 0 1 · · · 0...

......

......

0 0 0 0 · · · 1

.

The reduction continues until we reach the case where there are no rowsleft in Ei and Fi. But then there are no columns in the third vertical stripeand consequently no rows in the forth horizontal stripe and this goes on alldown to the bottom right and we see that in fact we are left with a matrixof the form [

Ci Di

]=[U1 V1

], (1.7)

since there can be no other surviving rows or columns. We proceed similarly:[U1 V1

]∼[0 1 U ′2 V ′20 0 U2 V2

]∼[0 1 0 00 0 U2 V2

]= [− ]⊕

[1 0 00 U2 V2

].

The allowed transformations for[U2 V2

]are the same than in (1.7) and

therefore we can again proceed iteratively, splitting off indecomposables

[− ],[

1 0],

[1 0 00 1 0

], . . . ,

1 0 · · · 0 00 1 · · · 0 0...

......

...0 0 · · · 1 0

.10


The corresponding indecomposables of the Kronecker problem are thereforeof the form

[z 1n | 1n z], for n ≥ 1. (1.8)

In each step the remaining matrix is of the form1 0 · · · 0 0 00 1 · · · 0 0 0...

......

......

0 0 · · · 1 0 00 0 · · · 0 Uj Vj

and we can proceed unless Uj has no column, which implies that the stripeabove has no rows, which in turn implies that the vertical stripe to the lefthas now rows and this argument runs through to the top and shows that infact we are left with the matrix [

Vj]

with coupled row and column transformation. We know from Exercise 1.1.1,that if the ground field K is algebraically closed, then the normal form isgiven by a Jordan matrix, which, in our language, decomposes as a directsum of Jordan blocks J(n, λ) = λ1n + J(n, 0), where λ ∈ K and J(n, 0)is the matrix having as only non-zero entries J(n, 0)ab = 1 if a = b− 1. Thecorresponding indecomposables of the Kronecker problem are of the form

[1n | J(n, λ)], for n ≥ 1. (1.9)

If K is not algebraically closed, normal forms still exist but are more com-plicated.

By collecting all indecomposables split off in the process, see (1.5), (1.6),(1.8) and (1.9), we get the solution of the Kronecker problem in case K isalgebraically closed:

Proposition 1.6. Suppose that the field K is algebraically closed. Then thefollowing is a complete list of indecomposables for the Kronecker problem[

1nz>

∣∣∣∣ z>1n], [1m | J(m,λ)], [J(m, 0) | 1m], [z 1n | 1n z], (1.10)

where n ≥ 0, m ≥ 1 are natural numbers, λ ∈ K, and z ∈ Kn denotes thezero column.

11

1 Matrix problems

Comment 1.7. Note that if K is not algebraically closed then we also get acomplete list if we substitute J(n, λ) by the appropriate indecomposables ofthe conjugation problem (which may be very difficult to get, as for exampleif K = Q). In the forthcoming we always will assume the field K to bealgebraically closed unless it is stated otherwise.

We have found an infinite but complete list of indecomposables, indexed bydiscrete values, namely m and n, and families indexed over the ground field,so-called one-parameter families.

Exercises

1.4.1 Let [A|B] be an element of the Kronecker problem with A,B ∈ K1×5. Noticethat [A|B] must decompose into some number t > 1 of smaller indecomposablesummands. What possible t might occur? Give for each possibility one example.

1.4.2 Consider those indecomposable elements [A|B] of the Kronecker problemwhich have square matrices A and B. How do those elements look like where notboth matrices A and B are invertible?

1.4.3 Proof that when two invertible matrices T and U satisfy

T

[1nz>

∣∣∣∣ z>1n]U−1 =

[1nz>

∣∣∣∣ z>1n],

where z is as in Proposition 1.6, then T = α1n+1 and U = α1n for some non-zeroscalar α ∈ K.

1.5 The three Kronecker problem

We consider finally three coupled matrices of the same size [A|B|C], that iswe consider the following matrix problem.

The 3-Kronecker problem is given by triples of matrices (A1, A2, A3)having the same size under the equivalence relation given by (A1, A2, A3) ∼(A′1, A

′2, A

′3) if there exist invertible matrices T,U such that A′i = TAiU

−1

for i = 1, 2, 3.

The straightforward generalization of n-tuples (A1, . . . , An) is called the n-Kronecker problem. The direct sum is defined in analogy to the case ofthe two subspace problem:

(A1, . . . , An)⊕ (B1, . . . , Bn) = (A1 ⊕B1, . . . , An ⊕Bn).

12

1.5 The three Kronecker problem

Again we write [A1|A2|A3] instead of (A1, A2, A3) and think of it as a bigmatrix with coupled columns. In this case our approach is very different.We first show the following result.

Proposition 1.8. For each λ, µ ∈ K define

Mλ,µ =

[10

∣∣∣∣ 01

∣∣∣∣ λµ].

Then the matrices Mλ,µ are pairwise non-equivalent indecomposables.

Proof. Suppose that Mλ,µ ∼ Mρ,σ, that is there are invertible matrices Uand T such that

T

[10

]U =

[10

], T

[01

]U =

[01

], and T

[λµ

]U =

[ρσ

]. (1.11)

Since the multiplication with U = [u] correspond to scalar multiplicationwith u, it follows from the first two equations that T12u = 12, and henceT = u−112. But then ρ = λ and σ = µ. In other words, the matrices Mλ,µ

are pairwise non-equivalent for different pairs (λ, µ).

They are also indecomposable: suppose that Mλ,µ ∼ [A|B|C] ⊕ [D|E|F ].But then, [D|E|F ] (or [A|B|C], the case is treated similarly) does not havecolumns and can only have rows. Since the first two columns of Mλ,µ arelinearly independent, the same holds for [A|B|C] and therefore [A|B|C] mustcontain two rows. Hence [D|E|F ] is the matrix of size 0× 0. Hence Mλ,µ isindecomposable.

Thus we found a two parameter family of pairwise non-equivalent in-decomposables. For a representation theorist, this is a point where it isclear that it is hopeless to try to write down a complete list of all inde-composables, since by such a list one would parametrize all modules overany finite-dimensional algebra simultaneously, see the end of Chapter 9 fora brief outline. This is why such a problem is called wild. However, if thesize of the matrix is fixed then it is possible to write down a complete listof representatives.

Exercises

1.5.1 For each λ = (a, b, c, d, e, f) ∈ K6 define the element Mλ of the 3-Kroneckerproblem as follows:

Mλ =

1 00 10 0

∣∣∣∣∣∣0 01 00 1

∣∣∣∣∣∣a bc de f

.13

1 Matrix problems

Determine when two such elements Mλ and Mµ are equivalent and when they arenot.

1.5.2 Use Exercise 1.4.3 to generalize the former exercise to get families of in-decomposable and pairwise non-equivalent elements of the 3-Kronecker problem,which are indexed with n(n+ 1) parameters.

1.6 Comments

We have encountered three different situations:

• A finite list of indecomposables as in the two subspace problem. Thisis the case of finite representation type.

• An infinite but complete list of indecomposables containing one (ormore) one-parameter family of pairwise non-equivalent indecompos-ables. This was the case in the Kronecker problem.

• A wild (hopeless) situation as in the 3-Kronecker problem. This is thecase of wild representation type.

These are the three representation types which occur in representationtheory studied here. The second and third case are those of infinite rep-resentation type. The first and second case together are called of tamerepresentation type.

Classifying is one of the important problems in representation theory (andthat is why we started with it), but clearly there are many others. We useda technique called “matrix problems”, which is very general and powerfuland has done the job quite satisfactorily. However you should have noticedif you tried to solve the three subspace problem, see Exercise 1.3.3, that themethod requires concentration and errors tend to be overseen and might slipthrough unnoticed.

Furthermore, we gained no structural insight and got just a plain list ofindecomposables. In the following chapters we will see that there are bettermethods, which are somehow “self-correcting”, which means that mistakestend to surface quickly. However, in order to be able to understand those“better” methods, we have to prepare first some theory. We start by rein-terpreting the considered problems from a new point of view.

As you may have noticed, we made no effort to formalize the term “matrixproblem”. This is due to the fact that we were mainly interested in some

14

1.6 Comments

key examples and the list produced by the techniques. Matrix problemis a very general concept allowing much broader context. For example,the classification of square matrices under transposition (A ∼ A′ if thereexists an invertible T with A′ = TAT>) is a nice and important matrixproblem, but it will not allow the translation into the language of quiversand representations as studied in the next chapter. Therefore we did notstress the formal definition but stuck to the examples.

15

1 Matrix problems

16

Chapter 2

Representations of quivers

All the examples of matrix problems we studied so far can be reformulatedin a diagrammatic way which yields a nice general point of view for lookingat matrix problems. This approach leads naturally to a more sophisticatedlanguage known as categories and functors and it will take large part of thischapter to develop these notions properly.

2.1 Quivers

Look again at the examples of Chapter 1. In the two subspace problem, weconsidered pairs of matrices (A,B) with the same number of rows under acertain equivalence relation. Identifying matrices with linear maps, we havethus considered diagrams

Kn′ Kn′′

Km@R A B

(2.1)

and were trying to find “good bases” for the involved vector spaces. Thedual problem (see Exercise 1.3.2) corresponds to diagrams of the form

Kn′ Kn′′

Km@I A B

.

In the Kronecker problem, we considered two matrices of the same size andin the three Kronecker problem three matrices of the same size under theequivalence relation of simultaneous row transformations and simultaneous

17

2 Representations of quivers

column transformations. Thus [A|B] and [A|B|C] corresponds to two andthree “parallel” linear maps, respectively:

Kl Km--A

Bresp. Kl Km.--

-AB

C

Again, the equivalence relation is given by arbitrary change of basis in thetwo vector spaces. Thus, these four matrix problems, are encoded by asimple diagram

, ,

r rr@R r rr@I r r-- orr r--- . (2.2)

The matrix problem may be recovered, by replacing the vertices by vec-tor spaces with basis, the arrows by linear maps (between the correspondingvector spaces) and considering two corresponding tuples of matrices as equiv-alent if one can be obtained from the other by change of basis. The diagramswe consider are therefore oriented graphs, where loops and multiple arrowsare explicitly allowed. We formalize this in the following.

A quiver is a quadruple Q = (Q0, Q1, s, t), where Q0 is the set of vertices,Q1 is the set of arrows and s, t are two maps Q1 → Q0, assigning thestarting vertex and the terminating vertex for each arrow. We also saythat α starts in s(α) and ends in t(α). A quiver Q is finite if Q0 and Q1

are finite sets.

Examples 2.1 (a) The quiver on the left in (2.2) is called two subspacequiver.

(b) The third quiver from the left in (2.2) is called Kronecker quiver.

(c) The quiver on the right in (2.2) is called three Kronecker quiver.More generally, the n-Kronecker quiver consists of two vertices 1, 2 andn > 1 arrows, which have 1 as starting vertex and 2 as terminating vertex.

♦

Usually, in examples, we will have Q0 = 1, . . . , n. For arrows α withs(α) = i and t(α) = j, we usually write. α : i→ j. In general, we denote byNQ0 the set of all functions Q0 → N, thus, if Q0 = 1, . . . , n then NQ0 = Nn.

18

2.2 Representations

To each quiver we can associate a matrix problem (the contrary is false, seeend of Section 2.2). Let Q be a finite quiver. Then the matrix problemassociated to Q is the pair (MQ,∼Q) where

MQ =⋃

d∈NQ0

MQ,d, MQ,d = (Mα)α∈Q1 |Mα ∈ Kdt(α)×ds(α)

and (Mα)α ∼Q (Nα)α if and only if there exists a family (Ui)i∈Q0 of invertiblematrices such that

Nα = Ut(α)MαU−1s(α) (2.3)

for each arrow α ∈ Q1.

For each M ∈MQ,d the vector d ∈ NQ0 is called the dimension vector ofM and shall be denoted by d = dimM .

Exercises

2.1.1 Draw the diagram of the quiver Q given by Q0 = 1, 2, 3, 4, 5, Q1 =α1, α2, α3, α4, s(αi) = i and t(αi) = i+ 1 for i = 1, . . . , 4.

2.1.2 Draw the quiver corresponding to the three subspace problem, described inExercise 1.3.3. This quiver is called the three subspace quiver.

2.1.3 Let Q be the quiver 1α−→ 2. Solve the corresponding matrix problem

(MQ,∼Q), that is, determine the indecomposables.

2.1.4 Determine the dimension vectors of the indecomposables in the Kroneckerproblem.

2.2 Representations

We will fix the ground field K and omit the dependence on K in our notationif no confusion can arise.

A representation of a quiver Q is a pair

V = ((Vi)i∈Q0 , (Vα)α∈Q1)

of two families: the first, indexed over the vertices of Q, is a family of finite-dimensional vector spaces and the second, indexed over the arrows of Q,consists of linear maps Vα : Vs(α) → Vt(α).

The zero representation, denoted by 0, is the unique family with Vi = 0(the zero vector space) for each i ∈ Q0.

19


It is common to write a representation “graphically” by replacing each ver-tex i by the vectorspace Vi and each arrow α : i → j by the linear mapVα : Vi → Vj . The dimension vector of a representation V is the vector(dimVi)i∈Q0 ∈ NQ0 .

Example 2.2. The indecomposable elements of the Kronecker problem withsquare matrices are [1m|J(m,λ)] for λ ∈ K and [J(m, 0)|1m]. They cor-respond to the following indecomposable representations of the Kroneckerquiver

Vm Vm--

X + λ

1resp. Vm Vm--

1

X ,

where Vm = K[X]/(Xm). ♦

Let V and W be two representations of a finite quiver Q. A morphismfrom V to W is a family of linear maps f = (fi : Vi →Wi)i∈Q0 such that foreach arrow α : i→ j we have

fjVα = Wαfi. (2.4)

We denote a morphism just like a function, that is, we write f : V → W toindicate that f is a morphism from V to W . Observe that equation (2.4)states that the following diagram commutes:

Vj

Vi

Wj

Wi

? ?-

-

fj

fi

Vα Wα

.

A morphism is an isomorphism if each fi is invertible and we say that Vand W are isomorphic representations if there exists an isomorphism fromV to W .

A basis of a representation V is a family (Bi)i∈Q0 where Bi is a basis of thespace Vi for each vertex i ∈ Q0. Each such basis yields a family of matricesV B = (V B

α )α∈Q1 where V Bα represents the linear map Vα in the bases Bs(α)

and Bt(α).

Proposition 2.3. Let V and W be two representations of a finite quiver Q.Then V is isomorphic to W if and only if for some (any) basis B of V andsome (any) basis C of W we have that V B and WC are equivalent elementsof the matrix problem associated to Q.

20

2.2 Representations

Proof. Notice that by chosing bases, we translate the linear invertible map fiinto an invertible matrix Ui. Condition (2.4) corresponds then to (2.3).

Let V and W be two representations of a quiver Q. The direct sum V ⊕Wis then defined as the representation given by the spaces (V ⊕W )i = Vi⊕Wi

and the linear maps (V ⊕W )α = Vα⊕Wα, which are defined componentwise.We denote V ⊕ V as a power by V 2 and inductively V i for lager exponentsi.

A representation V is indecomposable if and only if V 6= 0 and it is impos-sible to find an isomorphism V

∼−−→ V ′⊕V ′′ for any non-zero representationsV ′ and V ′′.

Proposition 2.4. Let V be a representation of a finite quiver Q. Then Vis indecomposable if and only if V B is indecomposable for some (any) basisB of V .

Proof. This an immediate consequence of the definitions and Proposi-tion 2.3.

We thus achieved a perfect translation. Solving one of the matrix problemsabove corresponds to classifying the indecomposable representations up toisomorphism. For instance, we get the following result.

Proposition 2.5. Each indecomposable representation of the quiver corre-sponding to the two subspace problem is isomorphic to precisely one repre-sentation of the following list.

, , , , ,

0 0

K@@R

K 0

0@@R

0 K

0@@R

K 0

K@@R [1]

0 K

K@@R [1]

K K

K@@R [1] [1]

.

Comments 2.6 (a) Observe that the strange matrices of the two subspaceproblem occurring in (1.3) correspond to natural representations.

(b) Notice that not every matrix problem we considered admits such astraightforward translation. For instance, in the coupled 4-block problem ofSection 1.4 we looked at quadruples of matrices[

C DE F

],

where the row and column transformations forD are coupled by conjugation.The corresponding quiver would look as follows (where we indicated theplaces of the matrices):

21


rr r@@R?

E

C

F

D

.

This quiver defines a wild case and does not correspond to our originalmatrix problem, since we have not expressed in our new language that wecan add rows from the lower stripe to the upper stripe nor that we can addcolumns from the left to the right stripe.

Exercises

2.2.1 Write the matrices given in Proposition 1.8 as representations of the corre-sponding quiver.

2.2.2 Use the spaces Vn and Vn+1 of Example 2.2 to write those indecomposablerepresentations of the Kronecker quiver which were not already given in the Exam-ple.

2.2.3 Decompose the following representation into indecomposables

K2

K2 K2

@@@R

[1 11 1

] [1 11 1

]

2.3 Categories and functors

We shall briefly explain the language of categories and functors, since itprovides a general language for the different concepts we shall encounter. Ifyou are already familiar with categories and functors you can skip all of thissection except the examples and read on in Section 2.4.

A category C is a class of objects (which we usually denote by the sameletter as the whole category) together with a familiy of sets C (x, y) whoseelements are called morphisms (one set for each pair of objects x, y ∈C ) together with a family of composition maps C (y, z) × C (x, y) →C (x, z), (g, f) 7→ g f (one for each triple of objects x, y, z ∈ C ) such thatfor each object x ∈ C there exists an identity morphism 1x ∈ C (x, x), thatis an element which satisfies 1x f = f and g 1x = g for any f ∈ C (w, x),

22


g ∈ C (x, y), any w, y ∈ C and such that the composition is associative, thatis (h g) f = h (g f) for any f ∈ C (w, x), g ∈ C (x, y), h ∈ C (y, z), anyw, x, y, z ∈ C .

This is a long definition! Intuitively, a category is something similar to whatyou obtain when you throw all sets and all maps between all these sets intoone big bag called Set, the category of sets.

To summarize: There are objects, which form a class; between any twoobjects there is a set of morphisms (possibly the empty set); morphismsmay be composed and the composition is associative; and there are identitymorphisms. We usually write f : x → y for a morphism f ∈ C (x, y) toremind us of the similarity with maps. We also often omit the compositionsymbol and write gf instead of g f .

Examples 2.7 (a) The category Set has as objects the class of all sets andas morphisms just all maps. The composition of morphisms in the categoryis just the composition of maps and the identity morphisms are the identitymaps.

(b) The category Vec has as objects K-vector spaces with the linear mapsas morphisms. The composition of morphisms and the identity morphismare again the obvious ones. The category vec has as objects the class of finite-dimensional K-vector spaces, again with the linear maps as morphisms.

(c) The category Top has the topological spaces as objects and continuousfunctions as morphisms. ♦

As you see you can take for the objects all representatives of a fixed alge-braic structure like topological spaces, rings, groups, abelian groups, finitelygenerated abelian groups and so on and so on. For the morphisms you takethe structure preserving maps between them, and for the composition justthe composition of maps. You always get a category. Let us look now atsome more and stranger categories.

Examples 2.8 (a) Let Q be a finite quiver. We will see that MQ can beviewed as a category. The class of objects is by definition just the set MQ

itself. If M,N ∈MQ then let

MQ(M,N) = (Ui)i∈Q0 | ∀α ∈ Q1, NαUs(α) = Ut(α)Mα.

23


Observe that the condition NαUs(α) = Ut(α)Mα is the same as (2.3) exceptthat we do not require the matrices Ui to be invertible.

It is easy to verify that MQ is indeed a category if the composition is givenby componentwise matrix multiplication and the identity morphisms are thetuples of identity matrices. However, morphisms are clearly not functionsbetween two sets in this example.

(b) The category repQ has as objects the representations of Q and as mor-phisms just the morphisms of representations. The composition is given bycomponentwise composition of linear functions and the identity morphismsare given by tuples of identity functions. Note, that as in the example before,morphisms are not given by a single function; in this case they consist of afamily of functions satisfying some compatibility property. If V and W arerepresentations of the quiver Q, we denote the morphism set repQ(V,W )also by HomQ(V,W ). ♦

In a category C a morphism f : x → y is called an isomorphism if thereexists a morphism g : y → x such that f g = 1y and g f = 1x. Twoobjects are said to be isomorphic in C if there exists an isomorphismbetween them.

Example 2.9. In the category MQ (see Example 2.8 (a)) two objects areisomorphic precisely when they are equivalent. So, the categorical conceptof isomorphism has just the right meaning we are interested in. ♦

The following concept will be used to relate different categories among eachother.

If C and D are categories then a covariant functor F : C → D associatesto each object x ∈ C an object of Fx ∈ D and to each morphism g ∈ C (x, y)a morphism Fg ∈ D(Fx, Fy) such that F1x = 1Fx for each object x ∈ Cand such that the composition is preserved, that is F (h g) = Fh Fg forany g ∈ C (x, y), h ∈ C (y, z), any x, y, z ∈ C . A contravariant functorF : C → D is very similar to a covariant functor except that it inverts thedirection of the morphisms, that is Fg ∈ D(Fy, Fx) for any g ∈ C (x, y) andconsequently F (h g) = Fg Fh for any g and h.

We shall meet many functors during the course of this book and limit our-selves here to two simple examples of covariant functors.

24


Examples 2.10 (a) Let Q be a finite quiver. Define a functor F : repQ→vec by FV =

⊕i∈Q0

Vi for any representation V of Q and Fg =⊕

i∈Q0gi

for any morphism of representations g.

(b) Let Q be a finite quiver. Then define a functor G : MQ → repQas follows: for an object M let (GM)i = Kdi , where d = dimM is thedimension vector, and (GM)α = Mα. Further, if U : M → N is a morphismin MQ, then define GU = U , with the abuse of notation that U denotes amatrix as well as the associated linear map in the canonical bases. ♦

If F : C → D and G : D → E are functors then we obtain a functorGF : C → E in the obvious way: (GF )x = G(Fx) for each object x and(GF )f = G(Ff) for each morphism f . The functor GF is called the com-position of F with G.

If C is a category, then the functor 1C : C → C defined by 1Cx = x and1C f = f for each object x and morphism f is called the identity functorof C .

Two functors F : C → D and G : D → C are called inverse to each otherif FG = 1D and GF = 1C . Note that F and G necessarily must have thesame variance, that is, they both must be covariant or both contravariant,see Exercise 2.3.4. If F and G are covariant, then they are called isomor-phisms and the categories C and D are called isomorphic. If F and G arecontravariant then they are called dualizations and the categories dual.

We will later see that it is rather seldom in practice that two categoriesare isomorphic, see Section 2.5, where we develop a weaker notion calledequivalence. The last piece of categorical terminology relates two functors.

If F : C → D and G : C → D are two covariant functors, then a morphismof functors (or natural transformation or just morphism) ϕ : F → G isa family (ϕx)x∈C of morphism ϕx ∈ D(Fx,Gx) such that ϕy Fh = Ghϕxfor any morphism h ∈ C (x, y).

If F : C → D is a given covariant functor, then the morphism F → F givenby the family (1Fx)x∈C is called identity morphism and will be denotedby 1F . A morphism ϕ : F → G of covariant functors is an isomorphism ifthere exists a morphism ψ : G→ F such that ψϕ = 1F and ϕψ = 1G.

25


Exercises

2.3.1 Prove that a morphism f : V → W of representations of a quiver is anisomorphism if and only if for each vertex i the linear map fi is bijective. Showthat in that case the family (f−1

i )i∈Q0of inverse maps constitutes an isomorphism

W → V of representations.

2.3.2 Prove a generalization of the previous exercise, namely, that a morphismϕ : F → G of functors F,G : C → D is an isomorphism if and only if for eachobject x ∈ C the morphism ϕx : Fx → Gx is an isomorphism. Show that in thatcase the family (ϕ−1

x )x∈C is an isomorphism of functors G→ F .

2.3.3 Verify carefully that all the properties stated in the definition of a categoryare satisfied in the two Examples 2.8.

2.3.4 Investigate when the functor GF is covariant and when it is contravariant,depending on the variance of F and G.

2.3.5 Let F : C → D and G : C → D be two contravariant functors. What is theapropriate condition for a family (ϕx)x∈C of morphisms ϕx ∈ D(Fx,Gx) to be amorphism of covariant functors?

2.4 The path category

A representation looks very much like a functor Q→ vec where Q is viewed“as a category” with vertices as objects and arrows as morphisms. Butof course this is nonsense, since there are no identity morphisms and nocomposition of arrows in Q. In the following we will enhance the quiver Qto a proper category. Therefore we will need the concept of paths in Q.

Let Q be a quiver (possibly infinite). A path of length l is a (l+ 2)-tuple

w = (j|αl, αl−1, . . . , α2, α1|i) (2.5)

where i, j ∈ Q0 and α1, . . . , αl ∈ Q1 such that s(α1) = i, t(αi) = s(αi+1) fori = 1, . . . , l − 1 and t(αl) = j.

We explicitly allow l = 0 but require then that j = i. The correspondingpath ei := (i||i) is called the identity path or trivial path in i. The lengthl of a path w is denoted by len(w).

We extend the functions s and t in the obvious way: s(w) = i and t(w) = jif w is the path (2.5). A path w of positive length l > 0 is called a cycle ifs(w) = t(w). Cycles are often also called oriented cycles in the literature.A cycle of length 1 is called loop.

26

2.4 The path category

The composition of two paths v = (i|αl, . . . , α1|h) and w = (j|βm, . . . , β1|i)is defined by

wv = (j|βm, . . . , β1, αl, . . . , α1|h).

Notice that we defined the composition of paths in the same order as func-tions, which is not at all standard in the literature, but rather up to the tasteof the author. A path (i|αl, . . . , α1|h) will often be denoted by αlαl−1 · · ·α1.

Let Q be a quiver (possibly infinite). The path category KQ of Q is thecategory whose objects are the vertices of Q and the morphisms from i toj form a vector space which has as basis the paths w with s(w) = i andt(w) = j. The composition is extended bilinearly from the composition ofpaths.

At this point we should pause a little and look at the curious fact that wedid not define the category of paths having as morphisms just the paths,as one might expect first. Indeed that would form a nice category also, butdue to reasons which shall become clear in the next chapter, we “linearize”the paths such that we can take sums and multiples.

A category is a K-category if its morphism sets are endowed with a K-vector space structure such that the composition is K-bilinear.

A functor F : C → D between K-categories is K-linear if C (x, y) →D(Fx, Fy), h 7→ Fh is K-linear for each pair of objects x, y ∈ C . If C is aK-category then mod C is the category of K-linear functors C → vec,that is, the objects of mod C are those functors and the morphisms arethe morphisms of functors with the obvious composition. For two functorsF,G ∈ mod C we write HomC (F,G) for the set of morphisms (mod C )(F,G).

Example 2.11. The path category KQ is a K-category. Moreover, eachrepresentation V of Q defines a K-linear (covariant) functor

V : KQ→ vec .

Conversely, any such functor gives rise to a representation of Q. ♦

In a K-category C the direct sum of two objects x and y is defined asobject z together with maps

x z y- - ιx

πx

ιy

πy

27


such that πxιx = idx, πyιy = idy, ιxπx + ιyπy = idz, πyιx = 0 and πxιy = 0.So, formally a direct sum is a quintuple (z, πx, πy, ιx, ιy). However, the objectz is –up to isomorphism– uniquely determined by x and y, see Exercise 2.4.6.This justifies the common abuse of language to call z itself the direct sumof x and y and denote it as x⊕ y.

Example 2.12. If C = repQ, the categorical direct sum corresponds tothe direct sum defined for representations above. Also in case C = MQ

the categorical direct sum corresponds to the direct sum in the language ofmatrix problems. ♦

Exercises

2.4.1 Verify that the morphisms of representations are precisely the morphismsbetween covariant K-linear functors KQ→ vec. Show that the category mod(KQ)is isomorphic to the category repQ.

2.4.2 If Q denotes the Kronecker quiver:

1 2 ,--α

β

then there are four morphism spaces in the category KQ. Determine the dimensionsof these spaces. Determine KQ(2, 1) as set. How many elements does it have?

2.4.3 Let Q be a finite quiver. Show that different objects of KQ are non-isomorphic.

2.4.4 For a finite quiver Q, prove that all morphism spaces in KQ are finite-dimensional if and only if there is no cycle in the quiver Q.

2.4.5 The adjacency matrix AQ of a quiver Q with vertices 1, . . . , n is the matrixof size n×n whose entry (AQ)ij is the number of arrows α ∈ Q1 with s(α) = j andt(α) = i. Prove that AQ is nilpotent (that is, there exists some positive integer tsuch that AtQ = 0) if and only if there is no cycle in Q. For this, show first, that for

each t, the entry (AtQ)ij equals the number of paths w of length t with s(w) = j,t(w) = i.

Conclude from this that in case Q has no cycle then AnQ = 0, where n is the

number of vertices. Furthermore show that the matrix B = 1n + AQ + A2Q +

. . .+ An−1Q measures the dimension of the morphism spaces in KQ, namely Bij =

dimK (KQ(j, i)).

2.4.6 Let C be a K-category. Suppose that the quintuples (z, πx, πy, ιx, ιy) and(z′, π′x, π

′y, ι′x, ι′y) are two direct sums of the objects x and y in C . Proof that z is

isomorphic to z′.

28

2.5 Equivalence of categories

2.5 Equivalence of categories

As we have seen there is a very close relationship between the category ofrepresentations repQ and the category of matrix problems MQ associatedto Q. In the following we would like to clarify this relationship completely.

We recall that two categories C and D are called isomorphic if there existtwo functors F : C → D and G : D → C such that GF = 1C and FG = 1D .

However, this notion is in most concrete cases far too restrictive. It is moreconvenient to look at some slight generalization: two categories C and Dare called equivalent if there exist two functors F : C → D and G : D → Csuch that GF ' 1C and FG ' 1D , that is, if there exists isomorphisms offunctors ϕ : GF → 1C and ψ : FG → 1D . In that case the functors F andG are called equivalences or quasi-inverse to each other.

Proposition 2.13. Let Q be a finite quiver. Then the categories repQ andMQ are equivalent.

Proof. We already have constructed the functor

F : MQ −→ repQ,

as application on the objects, see Example 2.10 (b). The definition on mor-phisms is straightforward.

To define a quasi-inverse G of F we choose a basis BV for each representationV . Define nV,i = dimVi for each representation V and each vertex i. Werecall that for each arrow α : i → j we get a matrix MV

α ∈ KnV,j×nV,i

representing the linear map Vα in the bases BVi and BV

j . Moreover, the

tuple G(V ) = (MVα )α∈Q1 defines an object of MQ. Furthermore, for each

morphism f : V → W of representations of Q we define G(f) = (Ufi )i∈Q0 ,

where Ufi is the matrix representing the linear map fi in the bases BVi and

BWi .

To see thatG : repQ −→MQ

is a functor we have to verify that G preserves identity morphisms and thecomposition. Indeed G(1V ) = 1G(V ) holds since the morphism 1V is the fam-ily of identity maps 1Vi : Vi → Vi which are expressed as identity matrices,since for both spaces we choose the same basis BV

i . For the composition,let U, V and W be representations of Q and f : U → V and g : V → W bemorphisms of representations. Then, for a vertex i of the quiver, G(f)i is

29


the matrix representing fi : Ui → Vi in the basis BUi and BV

i respectively.Similarly, G(g)i represents gi in the Basis BV

i and BWi respectively. There-

fore G(g f)i = G(g)iG(f)i holds for all i. This shows G(g f) = G(g)G(f)and finishes the proof that G is a functor.

It remains to see that F and G are quasi-inverse to each other. Furthermore,we denote by ψV,i : K

nV,i → Vi the linear map representing the identity ma-trix in the canonical basis and BV

i respectively. Note that KnV,i = FG(V )iholds by definition. To see that the family ψV = (ψV,i)i∈Q0 defines a mor-phism FG(V ) → V , we have to verify that for each arrow α : i → j thefollowing diagram commutes:

FG(V )i

FG(V )j

Vi

Vj

? ?-

-

FG(V )α Vα

ψV,j

ψV,i

.

We show this by looking at these maps in special bases: for Vi and Vj wechose the bases BV

i and BVj respectively and for FG(V )i and FG(V )j we

choose the canonical bases. The linear map Vα is then represented by thematrix G(V )α = MV

α , the maps ψV,i and ψV,j by identity matrices andFG(V )α also by G(V )α. This shows that ψV is a morphism of represen-tations. Since for each vertex ψV,i is bijective, it is an isomorphism, seeExercise 2.3.1.

Thus we have now a family (ψV )V ∈repQ of isomorphisms ψV : FG(V ) → Vof representations. To see that this family consitutes a morphism FG →1repQ of functors repQ → repQ it must be shown that for each morphismf : V → W of representations the following diagram on the left hand sidecommutes.

FG(V )

FG(W )

V

W

? ?-

-

FG(f) f

ψW

ψVFG(V )i

FG(W )i

Vi

Wi

? ?-

-

FG(f)i fi

ψW,i

ψV,i

By definition, this means that for each vertex i the diagram on the right handside commutes. Indeed, in the bases BV

i and BWi for Vi and Wi respectively

and the canonical bases for FG(V )i and FG(W )i, the linear maps fi andFG(f)i are represented by G(f)i, whereas ψV,i and ψW,i are represented byidentity matrices.

30

2.6 A new example

To see that ψ is an isomorphism of functors we have to give an inverse. Forthis we use Exercise 2.3.2 to see that for each representation V , the familyψ−1V = (ψ−1

V,i)i∈Q0 is an isomorphism of representations which is an inverseof ψ.

To get an isomorphism GF → 1MQwe could proceed very similarly. But

we will choose a much simpler way by restricting the choice of the bases foreach vector space of the form Kt to be always the canonical basis. It thenhappens that GF (M) = M for each M ∈ MQ. Thus GF = 1MQ

and ϕ isthe identity morphism.

We have intentionally written down all the details in the proof to show howeach categorical definition, which is involved, can be brought down in oursetting to statements about linear maps and matrices representing them.

Exercises

2.5.1 Let Q be the quiver which has a single vertex and no arrows. Show that vecand repQ are isomorphic categories. In this sense the study of representations ofquivers generalizes linear algebra of finite-dimensional vector spaces.

2.5.2 Let Mat be the category whose objects are the natural numbers (including 0)and whose morphism spaces Mat(n,m) are the sets Km×n of matrices of size m×nand entries in the field K. The composition in Mat is given matrix multiplication.Show that vec and Mat are equivalent categories.

2.6 A new example

We will consider a new class of problems starting from a family of quivers,which are called linearly oriented, and look as follows:

Q : r r r r r r- - - --1 2 3 n− 2 n− 1 n

α1 α2 αn−2 αn−1

We shall denote this quiver by−→An. The following result shows, that the

classification problem can be solved completely for−→An.

Theorem 2.14. Each indecomposable representation of−→An is isomorphic

to a representation

[j, i] : 0→ . . .→ 0→ K[1]−→ . . .

[1]−→ K → 0→ . . .→ 0,

31


where the first (that is, leftmost) occurrence of K happens in place j and thelast (that is, rightmost) in place i for some 1 ≤ j ≤ i ≤ n.

In particular,−→An is of finite representation type and there are n(n+1)

2 inde-composables, up to isomorphism.

Proof. Let V be an indecomposable representation of−→An. Let i be the

minimal index such that Vαi is not injective and set i = n if no such indexexists. Similarly, let j be the maximal index such that Vαj−1 is not surjectiveand set j = 1 if no such index exists. We shall show that V is isomorphicto [j, i].

If i < n then Vα1 , . . . , Vαi−1 are all injective, but Vαi is not. Then we letSi be a complement of Li = KerVαi , and set inductively Sh = V −1

αh(Sh+1),

Lh = V −1αh

(Lh+1) for h = i − 1, i − 2, . . . , 1. Note, that Sh ⊕ Lh = Vh forh = 1, . . . i. We thus see that V decomposes into

(L1 → . . .→ Li → 0→ . . .→ 0)⊕ (S1 → . . .→ Si → Vi+1 → . . .→ Vn)

and since V is indecomposable and Li 6= 0 the right summand must be zero.

Thus we have shown so far that Vh = 0 for h > i and that all maps Vαh areinjective for h < i. This implies that j ≤ i. We observe that if i = n thenall these statements are trivially true or void.

If j > 1 then Vαj , Vαj+1 , . . . , Vαi−1 are surjective and hence bijective, butVαj−1 is not. Let Rj be a complement of Mj = Vαj−1(Vj−1) and set in-ductively Mh = Vαh−1

(Mh−1), Rh = Vαh−1(Rh−1), for h = j + 1, . . . i. We

therefore conclude that V decomposes into

(0→ . . .→ 0→ Rj → . . .→ Ri → 0→ . . .→ 0)⊕(V1 → . . .→ Vj−1 →Mj → . . .Mi → 0→ . . .→ 0).

The indecomposability of V implies now that the latter one is zero, sinceRj 6= 0.

This shows that Vh = 0 for h < j and that Vαh is bijective for h = j, . . . , i−1.We observe that in case j = 1 all these statements are trivially true or void.Thus V is isomorphic to

0→ . . .→ 0→ Kd 1d−→ . . .1d−→ Kd → 0→ . . .→ 0,

where d denotes the dimension of the spaces Vj , . . . , Vi. But this represen-tation is isomorphic to the direct sum of d copies of [j, i]. By the indecom-posability of V it follows that d = 1 and that V is isomorphic to [j, i].

32

2.6 A new example

Thus, we have determined the objects of the category rep−→An: they are, up

to isomorphism, direct sums of the representations [j, i]. Now we turn our

attention to morphisms between representations of−→An. If V and W are

two representations, we first write them as direct sum of indecomposablerepresentations, say V '

⊕sa=1 Va and W '

⊕tb=1Wb, where each Va and

each Wb is of the form [j, i] for some 1 ≤ j ≤ i ≤ n. A morphism ϕ : V →W is then given by a matrix of morphisms ϕba : Va → Wb. Therefore,we are reduced to determine the morphisms between two indecomposablerepresentations [j, i] and [j′, i′].

Lemma 2.15. The morphism space Hom([j, i], [j′, i′]) is non-zero if andonly if j′ ≤ j ≤ i′ ≤ i. Moreover, in that case, Hom([j, i], [j′, i′]) is one-dimensional.

Proof. Suppose that ψ : [j, i] → [j′, i′] is a morphism. Clearly, there is al-ways the zero morphism with ψh = 0 for all h. If we suppose that ψ isnot identically zero then the two intervals [j, i] and [j′, i′] must have someintersection, that is, m = max(j, j′) ≤ min(i, i′) = M . For each h withm ≤ h ≤ M , the map ψh is just scalar multiplication with some factor λh.But if m ≤ h, h+ 1 ≤M , then the commutative square

K

K

K

K

? ?-

-

[1]

[1]

[λh] [λh+1]

shows that λh = λh+1. Hence, if there is a non-zero morphism then it is just

a non-zero scalar multiple of the morphism ι = ιj′,i′

j,i : [j, i] → [j′, i′], whereιh = [1] for each h = max(j, j′), . . . ,min(i, i′).

It remains to determine the condition, when a non-zero morphism ψ canexist. If j < j′ then we have a commuting square

0

K

K

K

? ?-

-

[1]

[1]

ψj′−1 ψj′

which shows that ψj′ = 0 and consequently ψ = 0. Similarly, the case i < i′

is excluded. Since m ≤ M we get j ≤ i′ and therefore j′ ≤ j ≤ i′ ≤ i is

33


a necessary condition for Hom([j, i], [j′, i′]) to be non-zero and in that casethis space is one-dimensional.

Hence we have

Hom([j, i], [j′, i′]) =

K ιj

′,i′

j,i , if j′ ≤ j ≤ i′ and j ≤ i′ ≤ i0, else.

Notice that the maps ιj′,i′

j,i behave multiplicatively, that is ιj′′,i′′

j′,i′ ιj′,i′

j,i = ιj′′,i′′

j,i .

We call a non-zero morphism between two indecomposable representationsof a quiver Q irreducible if it cannot be written as a sum of composi-tions, where each composition consists of two non-isomorphisms betweenindecomposables.

Hence from [j, i] there are, up to scalar multiples, at most two irreduciblemorphisms starting, namely ιj−1,i

j,i (if 1 < j) and ιj,i−1j,i (if j < i).

Putting everything together, we have the following diagram of irreduciblemaps between indecomposable representations in case n = 5.

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R[5, 5]

[4, 5]

[3, 5]

[2, 5]

[1, 5]

[4, 4]

[3, 4]

[2, 4]

[1, 4]

[3, 3]

[2, 3]

[1, 3]

[2, 2]

[1, 2]

[1, 1]

(2.6)

Notice that the whole diagram is commutative, i.e. each square in it iscommututaive. But there is still more structure inside of it which we shalldiscover in Section 6.3.

Exercises

2.6.1 Decompose the following representation V of−→An into indecomposables: Vi =

K2 for all i = 1, . . . , n and Vαi =

[0 10 0

]for all i = 1, . . . , n− 1.

2.6.2 Show that if f : V → W is a morphism of representations of a quiver Qthen Im f = (fi(Vi))i∈Q0

defines a subrepresentation of W , that is a family ofsubspaces W ′i ⊆ Wi such that Wα(W ′i ) ⊆ W ′j for each arrow α : i → j. Explainhow Wα defines (Im f)α.

34

2.6 A new example

2.6.3 Use the previous exercise to prove that an irreducible morphism f : V →Wbetween indecomposable representations satisfies that either all fi are injective orall fi are surjective.

35


36

Chapter 3

Algebras

Representations of quivers is a modern language which now will be connectedto the classic one of modules over an algebra. This translation will take upthe whole chapter. With this three different languages are developed for thesame thing, each of which with a distinctive flavour.

3.1 Definition and generalities about algebras

What do matrix problems and representations of quivers have to do with al-gebras and modules? We will need the whole chapter to answer the questionin a satisfactory way.

To fix notation, we start with the definition of an algebra and gradually seehow the previous two chapters link to the new concepts.

All rings will be unitary unless explicitly stated. If A is a ring, its center is

Z(A) = z ∈ A | az = za for all a ∈ A,

that is, Z(A) consists of those elements which commute with all elementsof the ring. Note that by definition Z(A) = A holds if and only if the ringA is commutative. Furthermore, it is easy to see that Z(A) is closed underaddition and multiplication, contains 0 and 1 and is closed under additiveinverses. Therefore Z(A) is a subring of A.

Example 3.1. The center of Kn×n is λ1n | λ ∈ K, see Exercise 3.1.1. ♦

An algebra over a commutative ring R, or R-algebra for short, is a ring A

37

3 Algebras

together with an inclusion ι : R→ Z(A), that is, an injective homomorphismof rings.

The inclusion ι induces the external multiplication

R×A→ A, (r, a) 7→ r · a = ι(r)a.

If R is a field then this implies that A is an R-vector space and thus has awell defined dimension over R. In most cases considered in this book, thisdimension will be finite, but exceptions will occur also.

A homomorphism of R-algebras (A, ιA) → (B, ιB) is a ring-homomor-phism f : A→ B such that ιB = fιA. It is common to write just f : A→ Bin that case and to omit the information on the inclusions if no confusioncan arise. A homomorphism of R-algebras f is called an isomorphism if itis invertible: that is, there exists g : B → A such that gf = 1A and fg = 1B.

In most cases we will describe an algebra over a field K by giving a K-vectorspace A together with a multiplication which has a unit 1A. The inclusionι is then meant to be λ 7→ λ1A.

Let A be an K-algebra, where K is a field. Since A is in particular aring, we can consider left and right A-modules and in fact these will bethe A-modules, even if A is considered as algebra. If M is a left A-moduleand we restrict the multiplication A×M → M, (a,m) 7→ am to K ×M →M, (λ,m) 7→ ι(λ)m, we see that M is a K-vector space. Hence the dimensionof a module M is well defined and denoted by dimM . In this book, we willstudy such modules as well as the algebras themselves.

Let A be a finite-dimensional K-algebra and M an A-module. A submod-ule U of M is by definition an abelian subgroup U ⊆M such that au ∈ Ufor all a ∈ A and all u ∈ U . If we restrict A to K we see that U is asubspace of M . The multiplication in M induces naturally a multiplicationin the quotient space M/U by a(m+U) := am+U . Hence each submoduleU ⊆M yields a quotient module M/U . Each K-algebra admits the zerospace as module, which is called the zero module and denoted by 0.

Given a K-algebra A and two left (resp. right) A-modules M and N .Then each homomorphism f : M → N of left (resp. right) A-modules,that is, each homomorphims of abelian groups f : M → N which satis-fies af(m) = f(am) for all a ∈ A and all m ∈ M is automatically K-linear. Therefore the set of homomorphisms HomA(M,N) is a K-vectorspace. If both modules are finite-dimensional then also HomA(M,N) isfinite-dimensional. We usually consider left modules. The category of finite-dimensional left A-modules will be denoted by modA.

38

3.1 Definition and generalities about algebras

If M = N , we also denote EndA(M) = HomA(M,M) and call its elementsendomorphisms of M . The space EndA(M) is clearly closed under sumsand compositions. It contains the zero map and the identity map idM as theneutral element of the sum and the composition respectively. Furthermore,these two binary operations satisfy the distribution laws. Also, the multiplesλidM with λ ∈ K, commute with any other endomorphism. Hence EndA(M)is a K-algebra, called the endomorphism algebra of M .

If R and S are rings then a R-S-bimodule M is a left R-module whichis also a right S-module such that (rm)s = r(ms) for all r ∈ R, s ∈ Sand m ∈ M . If R and S are both K-algebras, then it follows that λm =(1Rλ)m = m(λ1S) = mλ for all λ ∈ K.

Exercises

3.1.1 Prove that Z(Kn×n) = λ1n | λ ∈ K.

3.1.2 Show that a homomorphism of R-algebras is an isomorphism if and only ifit is bijective.

3.1.3 Show that the ring L of lower triangular matrices in Kn×n is a K-algebra;determine Z(L), the inclusion K → Z(L) and the dimension of L over K.

3.1.4 Show that, if A is an R-algebra, then the set r1A | r ∈ R is contained inZ(A).

3.1.5 Explain why complex conjugation C→ C, a+ bi 7→ a− bi is an isomorphismof R-algebras, but not even a homomorphism of C-algebras.

3.1.6 Let A be an R-algebra. Prove that any A-module M is also an R-module.In particular, if R is a field then M is an R-vector space.

3.1.7 Prove that there are, up to isomorphism, precisely two C-algebras of dimen-sion 2 over C, namely C× C and C[X]/(X2).

3.1.8 Let R be any ring. Observe that R can be considered as left R-module. Theendomorphism ring EndR(R) consists then of all homomorphisms of abeliangroups ϕ : R → R which satisfy ϕ(rx) = rϕ(x) for all r, x ∈ R. Prove thatf : EndR(R) → Rop, ϕ 7→ ϕ(1R) is an isomorphism of rings, where Rop is theopposite ring of R, that is Rop = R as abelian groups and the multiplication ∗ inRop is given by r ∗ s = s · r, where the latter is the multiplication in R. Investigatewhat happens if R is considered as a right R-module. Prove then that EndR(R) isisomorphic to R.

39

3 Algebras

3.2 The path algebra

We now want to associate an algebra AQ, called path algebra, to a finitequiver Q. Recall that KQ denotes the path category, as explained in Sec-tion 2.4. In particular the objects of KQ form the set Q0. As vector space,the algebra AQ is defined as

AQ =⊕

i,j∈KQKQ(i, j).

The multiplication in AQ is obtained by extending the composition of pathsbilinearly, setting vw = 0 whenever s(v) 6= t(w).

Then we have that AQ is an associative ring with unit 1AQ =∑

i∈KQ(i||i) =∑i∈KQ ei. Moreover, the ground field acts centrally, that is, for λ ∈ K and

a ∈ AQ, we always have (λ1AQ)a = a(λ1AQ). Thus, AQ is a K-algebra.

It is common to denote the path algebra AQ as KQ, exactly like the pathcategory. Note that AQ in general is not commutative. For example A−→A 2

isnot commutative.

Examples 3.2 (a) Notice, that in A−→An, there is a path from i to j if and

only if i ≤ j. Hence the algebra A−→Anis isomorphic to the lower triangular

matrices of size n× n, under the mapping induced by

(j|αj−1, . . . αi|i) 7→ E(ji),

where E(ji) is the n × n-matrix, whose unique non-zero entry equals 1 andis located in the j-th row and i-th column.

(b) If Q is the Kronecker quiver, see Example 2.1 (b), then the algebra AQis called the Kronecker algebra. Note that it has dimension 4 over thefield K. ♦

We now can start to relate modules over the path algebra AQ with therepresentations of the quiver Q.

Theorem 3.3. For any finite quiver Q, the category modAQ of finite-dimensional left modules over the path algebra AQ is isomorphic to the cat-egory repQ of representations of Q.

40

3.2 The path algebra

Proof. Given any representation V of Q, we get a left AQ-module

V ′ =⊕i∈KQ

Vi

by defining the multiplication V ′w with a path w = (j|αl, . . . α1|i) on a family(vh)h∈KQ as the family having Vαl . . . Vα1(vi) in the j-th coordinate andzero elsewhere. Notice that V ′ is finite-dimensional, since Q is finite and bydefinition a representation has finite-dimensional vector spaces attached toeach vertex. Conversely, given a finite-dimensional left AQ-module M , wedefine Mi = eiM = eim | m ∈ M. We have then M =

⊕i∈KQMi and

can easily define a representation by setting Mα : Mi →Mj , eim 7→ (j|α|i)mfor any arrow α : i→ j in Q.

If ϕ : V → W is a morphism of representations, then we define ϕ′ =⊕i∈KQ ϕi : V

′ → W ′, which is a homomorphisms of AQ-modules. Con-versely, any homomorphism ψ : M → N of finite-dimensional AQ-modulesgives rise to a morphism of representations (ψi)i∈Q0 , where ψi : Mi → Ni,eim 7→ ψ(eim) = eiψ(m).

Remark 3.4. Since the category modAQ is isomorphic to repQ, the cate-gorical concepts of direct sum and indecomposablity translate from repQ tomodAQ. Explicitly, the direct sum of two modules M and N is given asM ⊕ N as vector space with componentwise multiplication and a non-zeromodule M is indecomposable if M ' N ⊕ N ′ implies N = 0 or N ′ = 0.As with representation we use the notation Mn to denote the direct sumM ⊕ . . .⊕M with n summands. ♦

Example 3.5. The algebra A of lower triangular matrices of size 5 × 5 isisomorphic to A−→A 5

. Hence by Theorem 3.3 and 2.14, there are 15 indecom-posable A-modules up to isomorphism. ♦

Exercises

3.2.1 Let Q be the quiver with one vertex and one arrow. Notice that this arrownecessarily must be a loop. Prove that the path algebra AQ is isomorphic to thepolynomial algebra in one variable K[X].

3.2.2 Let A be the two subspace algebra, that is, the path algebra of the twosubspace quiver, see Example 2.1 (a). Show that the algebra A is isomorphic tothe algebra B, which consist of the matrices M ∈ K3×3 such that M12 = M13 =M31 = M32 = 0 with the usual matrix multiplication as multiplication in B. Givean explicit isomorphism, which maps each of the 5 paths in A to a matrix of theform E(ji) for some i, j.

41

3 Algebras

3.2.3 Determine the number of isomorphism classes of indecomposable A-modulesof a fixed dimension d if A is the Kronecker algebra.

3.3 Quotients by ideals

We will show that to study the module categories of finite-dimensional alge-bras it is enough to consider quotients of path algebras, see Theorem 3.28.It will be useful to see such quotients right from the beginning in order tohave concrete examples to illustrate the forthcoming theory.

A left (resp. right) ideal I of a K-algebra A is a subset of A, which definesa subgroup with respect to the addition such that ai ∈ I (resp. ia ∈ I) foreach a ∈ A and each i ∈ I. A subset which is both a left and right ideal iscalled both-sided ideal, or just ideal for short. It follows straightforwardfrom the definitions that each left (resp. right) ideal is always a K-subspaceof A.

If I is an ideal of A, the set A/I is an abelian group as quotient of abeliangroups. The multiplication in A induces a well-defined multiplication inA/I, precisely because of the ideal property.

Example 3.6. Let Q be the quiver with one vertex and one arrow α, whichthen must be a loop, see Exercise 3.2.1. Then let I ⊆ AQ be the idealgenerated by the αn. Then the algebra AQ/I is isomorphic to the algebraK[X]/(Xn). ♦

A left (resp. right) ideal of a category C is a family (I (x, y))x,y∈C ofsubsets I (x, y) ⊆ C (x, y) such that for each j ∈ I (x, y), each object z ∈ Cand each c ∈ C (y, z), (resp. for each t ∈ C and each d ∈ C (t, x)) thecomposition cj belongs to I (x, z) (resp. jd belongs to I (t, y)). If thecategory C is a K-category, the ideal is assumed to consist of subspacesI (x, y) ⊆ C (x, y) and not just subsets. Similary a (both-sided) ideal ofa category C is a left ideal which is also a right ideal.

Example 3.7. Let Q =−→An be the linearly oriented quiver with n vertices

and I = (I (j, i))ni,j=1 be the family given by I (j, i) = K−→An(j, i) if j−i > 2

and I (j, i) = 0 else. Then I is an ideal of K−→An. ♦

Whenever we have an ideal I of a K-category C we may consider the quo-tient category C /I , which has the same objects than C , and morphismspaces (C /I )(x, y) = C (x, y)/I (x, y). That the composition of C /I ,

42

3.4 Idempotents

which is induced by the composition of C , is well defined is due to the idealproperty, as in the theory of rings, see Exercise 3.3.2.

Exercises

3.3.1 Determine K−→An/I if I is the ideal of K

−→An as defined in Example 3.7.

3.3.2 Let I be an ideal of a category C . Verify that the composition in thequotient category C /I is well defined.

3.3.3 Let A and B be two algebras and f : A → B an algebra homomorphism.Prove that Ker f = a ∈ A | f(a) = 0 is always an ideal of A.

3.3.4 Let Q be a quiver with one vertex and two arrows α, β, which must be loops.Observe that there is an obvious algebra homomorphism π : AQ → K[X,Y ], suchthat π(α) = X and π(β) = Y . Show that π is surjective and determine its kernelπ−1(0).

3.3.5 Let C be a K-category with finitely many objects and finite-dimensionalmorphism spaces and let AC =

⊕x,y∈C C (x, y). Show that AC is an algebra and

that modAC and mod C are isomorphic categories.

3.4 Idempotents

Not any K-algebra is isomorphic to a path algebra of some quiver. However,we will show that to any algebra A we may associate a quiver Q which willbe of central importance. First, we give the general idea of how to work ourway back from the algebra to a category, and then in a second step back to aquiver. We first outline this construction in the example of lower triangularmatrices.

Example 3.8. Let A be the K-algebra of lower triangular n× n-matrices.We constructed this algebra as path algebra of the linearly oriented quiver−→An and would like to see how this quiver can be reconstructed from thealgebra A. Which elements of A do correspond to the vertices and to the

arrows of−→An? We have seen in Example 3.2 (a), that the elements E(ii) for

i = 1, . . . , n are related to the vertices whereas E(i+1 i) for i = 1, . . . , n−1 arerelated to the arrows. We have now to describe these elements in terms of themultiplication: the elements e = E(ii) satisfy e2 = e, whereas the elementsy = E(i+1 i) are vaguely those which cannot be obtained as products in A“without involving” some element E(jj). ♦

43

3 Algebras

An element e of an algebra A is called idempotent if e2 = e. Note that 0and 1 are always idempotents. If e is an idempotent then also f = 1 − e,since f2 = 1−2e−e2 = 1−2e+e = f . Furthermore, these two idempotentssatisfy ef = 0 = fe and e + f = 1. In general, two idempotents e and e′

are said to be orthogonal if ee′ = 0 = e′e. A non-zero idempotent e isprimitive if for any two orthogonal idempotents e′ and e′′ with e′ + e′′ = ewe must have e′ = 0 or e′′ = 0. A set e1, . . . , et of pairwise orthogonalidempotents is called complete if

∑ti=1 ei = 1.

Proposition 3.9. In each finite-dimensional K-algebra A there exists afinite, complete set of pairwise orthogonal, primitive idempotents.

Proof. If 1 is primitive the set is 1 and we are done. Otherwise, thereexist two non-zero, orthogonal idempotents e′ and e′′ such that 1 = e′ + e′′.We set L1 = 1 and L2 = e′, e′′ and construct for m ≥ 3 iteratively setsLm which consist of m pairwise orthogonal, non-zero idempotents such that∑

x∈Lm x = 1. This iteration continues as long as we find an idempotentx ∈ Lm which is not primitive: we then find two non-zero, orthogonalidempotents y′ and y′′ such that x = y′ + y′′. Note that, xy′ = y′2 + y′′y′ =y′ = y′x and therefore y′ is orthogonal to any z ∈ Lm \ x since y′z =y′xz = 0 and similarly zy′ = 0. We then set Lm+1 = Lm \ x ∪ y′, y′′.It remains to show that there exists some integer M for which each idempo-tent of LM is primitive. For this, we consider the sets xA = xa | a ∈ A foreach x ∈ LM . Note that xA is a K-vector space, that xA 6= 0 since x 6= 0 iscontained in xA and that A =

⊕x∈LM xA as vector spaces. The latter holds

since a = 1a =∑

x∈LM xa ∈∑

x∈LM xA for each element a ∈ A and that

for each x ∈ LM we have xA ∩(∑

z 6=x zA)

= 0 since∑

z 6=x zA = (1 − x)A

and the idempotents x, 1− x are orthogonal.

Thus for each m, we have that dimK A =∑

x∈Lm dimK xA. Since the spacesxA are non-zero and the dimension of A is finite, we must have m ≤ dimK A.This finishes the proof.

Remark 3.10. Note that the proof shows that each complete set of pairwiseorthogonal, primitive idempotents is finite. ♦

Once we have found such a complete set S of pairwise orthogonal, primitiveidempotents we can define a category AS : the objects form the set S andthe morphism sets are given by AS(x, y) = yAx = yax | a ∈ A and thecomposition is induced by the multiplication of A. The category AS is aK-category.

44

3.4 Idempotents

Proposition 3.11. Let A be a finite-dimensional K-algebra and S a com-plete set of pairwise orthogonal, primitive idempotents. Then the categorymodA of finite-dimensional left A-modules is equivalent to the categorymodAS of K-linear functors AS → vec.

Proof. For each finite-dimensional left A-module M , we can define a K-linear functor Φ(M) : AS → vec by Φ(M)(x) = xM and Φ(M)(a) : M(x)→M(y),m 7→ am for each a ∈ AS(x, y). If f : M → N is a homomorphismof A-modules, then f(xM) = xf(M) ⊆ xN and therefore f induces theK-linear maps fx : xM → xN . The family (fx)x∈S satisfies fyΦ(M)(a) =Φ(N)(a)fx for each a ∈ AS(x, y), that is, (fx)x∈S is a morphism of functorsΦ(f) : Φ(M)→ Φ(N).

Note that A =⊕

x,y∈S AS(x, y). Therefore an element a ∈ A can be writtenin unique way a =

∑x,y∈S ax,y with ax,y ∈ AS(x, y) for all x, y ∈ S. Hence,

if F : AS → vec is a K-linear functor, then Ψ(F ) =⊕

x∈S F (x) is an A-module via the multiplication a·(mx)x∈S =

(∑x∈S ay,xmx

)y∈S . A morphism

g : F → G gives rise to a homomorphism Ψ(g) =⊕

x∈S gx : Ψ(M)→ Ψ(N).It is easy to see that Φ and Ψ are inverse to each other.

It will be convenient to have a different characterization for the primitivenessof an idempotent. A finite-dimensional algebra A is called local if eachelement a ∈ A is either invertible or nilpotent, that is, there exists apositive natural number ` such that a` = 0. We first need an auxiliaryresult concerning endomorphisms of an A-module.

Lemma 3.12. Let A a K-algebra and M a finite-dimensional A-module.Then for each endomorphism ϕ : M → M there exists a natural numbern ≥ 1 such that M = Imϕn ⊕Kerϕn.

Proof. The spaces ϕm(M) form a chain

M = ϕ0(M) ⊇ ϕ1(M) ⊇ ϕ2(M) ⊇ . . .

with decreasing dimensions. Therefore there exists an index n such thatϕn(M) = ϕn+1(M). Consequently ϕn(M) = ϕ`(M) for all ` ≥ n, in particu-lar for ` = 2n. Denote ψ = ϕn. Then for each m ∈M there exists an elementm′ ∈ M such that ψ(m) = ψ2(m′). Consequently m − ψ(m′) ∈ Kerψ andhence m = ψ(m′) + (m− ψ(m′)) ∈ Imψ + Kerψ. Thus M = Imψ + Kerψ.If m ∈ Imψ ∩ Kerψ then m = ψ(m′) and therefore 0 = ψ(m) = ψ2(m′).But ψ induces a bijection ψ(M) → ψ2(M) = ψ(M). Therefore ψ2(m′) = 0

45

3 Algebras

implies m = ψ(m′) = 0. This shows that M = Imψ ⊕ Kerψ as vectorspaces. The same decomposition holds as A-modules since both Imψ andKerψ are A-submodules. This concludes the proof of the assertion.

Proposition 3.13. Let A be a K-algebra. A finite-dimensional left (resp.right) A-module M is indecomposable if and only if EndA(M) is local.

Proof. Assume M is an indecomposable left A-module and ϕ ∈ EndA(M).Then, by Lemma 3.12, there exists an integer ` > 0 such that M =Imϕ` ⊕ Kerϕ`. Since M is indecomposable either Kerϕ` = M , that is, ϕis nilpotent, or Imϕ` = M , that is, ϕ` is an isomorphism. Since EndA(M)is a vector space of finite-dimension, the latter can only happen if ϕ itself isan isomorphism, hence invertible. This shows that EndA(M) is local.

Assume now that M is not indecomposable. Then there exist M1,M2 6= 0such that M = M1 ⊕M2. Denote by π1 : M →M1 the canonical projectionand by ι1 : M1 → M the canonical inclusion. Then ϕ = ι1π1 ∈ EndA(M)is idempotent, neither invertible nor nilpotent since ϕ 6= 0 and ϕ 6= idM .Hence EndA(M) is not local.

The proof for right A-modules is completely similar.

Proposition 3.14. A finite-dimensional algebra A is local if and only if 1Ais primitive if and only if A is indecomposable as left (resp. right) A-module.

Proof. If we consider A as right A-module, we know from Proposition 3.13,that A is indecomposable if and only if EndA(A) is local. Using Exer-cise 3.1.8, we see that EndA(A) is isomorphic to A. If we consider A as leftA-module, then we see that A is indecomposable if and only if Aop is local.It follows from the definition of being local that A is local if and only if Aop

is local. It remains to see that these conditions are equivalent to 1A beingprimitive.

First suppose that 1 = 1A is primitive. Assume that we have a decomposi-tion of A as left A-module, say A = M1 ⊕M2, then we denote for i = 1, 2by πi : A→Mi the canonical projections, and by ιi : Mi → A the canonicalinclusions. Then 1 = f1 + f2, where fi = ιiπi(1) for i = 1, 2. Note that fiare idempotents f2

i = ιiπi(1)ιiπi(1) = ιiπi(ιiπi(1)1) = fi, where the secondequation follows from the fact that πi and ιi are homomorphisms of leftA-modules and the last equation follows from πiιi = idMi . Now, since 1 isprimitive either f1 = 0 or f2 = 0 and consequently M1 = 0 or M2 = 0. Thisshows that A is indecomposable as left A-module.

46

3.5 Morita equivalence

Now suppose that 1 = f1 + f2 where f1 and f2 are non-zero orthogonalidempotents. Then A = A1 ⊕ A2, where Ai = Afi are left A-modules,which are non-zero since fi ∈ Ai. Hence A is not indecomposable as leftA-module.

Corollary 3.15. An idempotent e of a finite-dimensional algebra A is prim-itive if and only if the algebra eAe = eae | a ∈ A is local and this happensif and only if eAe is indecomposable as left (or right) eAe-module.

Proof. This follows immediately from Proposition 3.14 since e = 1B forB = eAe.

Exercises

3.4.1 Let B = K2×2 and A = B2×2. Determine a complete set of pairwise orthog-onal, primitive idempotents for the K-algebra A.

3.4.2 Show that a finite-dimensional algebra is local if and only if the nilpotentelements form an ideal. The latter is the general definition for a ring R to be local.It is equivalent to each of the following conditions: (i) there exists a unique maximalleft ideal; (ii) there exists a unique maximal right ideal; (iii) 0 6= 1 and for eachx ∈ R either x or 1− x is invertible.

3.4.3 Show that for any positive integer n the algebra K[X]/(Xn) is local.

3.4.4 Show that for n > 1 the algebra Kn×n is not local.

3.4.5 Let A = Kn×n and let S = e1, . . . , en where ei = E(ii). Show that anytwo idempotents ei and ej are isomorphic in the category AS .

3.5 Morita equivalence

If A is a finite-dimensional algebra, then the elements of a complete set S ofpairwise orthogonal, primitive idempotents of A are good candidates for thevertices of the quiver of A. Note that for each quiver Q, different verticesare non-isomorphic as objects of the category KQ, see Exercise 2.4.3. Inthe category AS , this must not be the case, see Exercise 3.4.5.

An algebra A is called basic if there exists a complete set S of pairwiseorthogonal, primitive idempotents, which are pairwise not isomorphic asobjects of AS .

47

3 Algebras

It seems that the idea of finding such a complete set S of pairwise orthogonal,primitive idempotents was not successful after all. We have no chance ofassigning a quiver Q to A such that the vertices of Q are S. Fortunately,there is an elegant way around that problem if we are primarily interested inthe module categories. This motivates the following definition: two finite-dimensional algebras A and B are said to be Morita equivalent, if modAand modB are equivalent.

Proposition 3.16. For any finite-dimensional algebra A there exists a basicfinite-dimensional algebra B which is Morita equivalent to A.

Proof. By Proposition 3.9, there exists a finite, complete set S of pairwiseorthogonal, primitive idempotents of A. We will show by induction on thenumber of elements of S that there exists a basic algebra B which is Moritaequivalent to A. Note that the statement is clear if S contains only oneelement.

If the elements of S are pairwise non-isomorphic, then A is already basicand we are done setting B = A. Otherwise, there exist e, f ∈ S which areisomorphic in AS , that is, there exist u ∈ AS(e, f) and v ∈ AS(f, e) suchthat uv = 1f = f and vu = 1e = e. We then define A′ = (1A−f)A(1A−f).Note that S′ = S \ f is a complete set of pairwise orthogonal, primitiveidempotents forA′. Thus by the induction hypothesisA′ is Morita equivalentto a basic algebra B.

All what remains to do is to show that A and A′ are Morita equivalent, thatis, we have to show that modA and modA′ are equivalent. By Proposi-tion 3.11 it is enough to show that the categories modAS and modA′S′ areequivalent.

We define the functor Φ: modAS → modA′S′ just as the obvious restriction.To define Ψ: modA′S′ → modAS we first observe that for any M ∈ modAS ,the map M(u) : M(e) → M(f) is bijective since it has an inverse, namelyM(v). Therefore we may define Ψ(M) : AS → vec on the objects by settingΨ(M)(f) = M(e) and Ψ(M)(x) = M(x) for all x 6= f . On homomorphisms,Ψ(M) is defined as follows:

Ψ(M)(β) =

M(β), if s(β) 6= f 6= t(β),

M(βv), if s(β) = f 6= t(β),

M(uβ), if s(β) 6= f = t(β),

M(uβv), if s(β) = f = t(β).

48

3.6 The radical

Then for any N ∈ modAS there exists a natural isomorphism ϕN : N →ΨΦ(N) which is given by ϕf = v and ϕx = 1N(x) for all x 6= f . Theseisomorphisms ϕN form a family which gives rise to an isomorphism of func-tors 1modAS → ΨΦ. It is easy to see that ΦΨ = 1modA′

S′. This finishes the

proof.

Exercises

3.5.1 Prove that Kn×n is Morita equivalent to K for all n ≥ 1.

3.6 The radical

We recall that if A is a finite-dimensional, local algebra then the nilpotentelements form an ideal radA, which is maximal as left and right ideal, seeExercise 3.4.2. The quotient A/ radA is then a skew field extension of K.The ideal is called Jacobson radical of A or just radical of A.

The radical plays a fundamental role in the forthcoming and therefore wecarefully outline these concepts here. First we generalize this notion toarbitrary algebras. The Jacobson radical radA of an algebra A is theintersection of all maximal left ideals.

Lemma 3.17. Let A be a finite-dimensional algebra. Then an element x ofA belongs to radA if and only if 1− ax is invertible for each a ∈ A.

Proof. Assume first that x belongs to radA. Let a ∈ A be any element. If1 − ax is not invertible, then the left ideal generated by 1 − ax is properlycontained in A. Since A is finite-dimensional, there exists a maximal leftideal I ⊂ A such that 1 − ax ∈ I. By definition x ∈ I and hence ax ∈ I.Consequently 1 = (1− ax) + ax ∈ I, which implies I = A, a contradiction.

Assume now, that 1 − ax is invertible for each a ∈ A. Let I ⊂ A be somemaximal left ideal of A and π : A → A/I the canonical projection. Since Iis maximal A/I is a field. If x 6∈ I, then π(x) 6= 0 and therefore there existsan inverse π(a), that is, π(ax) = π(1) and hence 1− ax ∈ I, which again isa contradiction.

Proposition 3.18. The radical radA of a finite-dimensional algebra A is aright ideal and equals the intersection of all maximal right ideals of A. Alsox ∈ radA if and only if 1− xa is invertible for all a ∈ A.

49

3 Algebras

Proof. We denote by R′ the intersection of all maximal right ideals of A.With a similar argument as given in the proof of Lemma 3.17, we see thatx ∈ R′ if and only if 1− xa is invertible for all a ∈ A.

Let x ∈ radA. We have to show that xa ∈ radA for each a ∈ A. Indeed,if xa 6∈ radA for some a ∈ A, then by Lemma 3.17 there exists an elementb ∈ A such that 1 − bxa is not invertible. Consequently, there exists amaximal left ideal I ⊂ A such that 1−bxa ∈ I. Since z = bx ∈ radA, 1−azis invertible and hence (1 − az)a = a(1 − za) ∈ I implies that a ∈ I andthen 1 = (1− za) + za ∈ I, a contradiction.

Therefore each x ∈ radA satisfies that 1 − xa is invertible for all a ∈ A.This shows radA ⊆ R′. By a similar argument, R′ is a left ideal andR′ ⊆ radA.

Let A be a finite-dimensional algebra and M a left A-module. Then theradical of M is defined as radM = (radA)M , that is, radM consists of allelements of M which can be written as finite sums

∑ti=1 rimi with ri ∈ radA

and mi ∈M .

Remark 3.19. It follows from the definition that rad(M ⊕N) = (radM)⊕(radN) for two A-modules M and N and that f(radM) ⊆ radN for eachhomomorphism f : M → N . ♦

Proposition 3.20 (Nakayama’s Lemma). If A is a finite-dimensional alge-bra and M 6= 0 a finite-dimensional left A-module, then radM is a propersubmodule of M .

Proof. Let S be a subset of M . We say that S generates M or that S isa generating set for M , if the smallest submodule of M which contains Sis M itself. This is equivalent to the fact that each element of M can bewritten as sum

∑ti=1 aisi with ai ∈ A and si ∈ S for some t ≥ 1.

Since M is finite-dimensional, any K-basis is a generating set for M . So, letnow S be a generating set with minimal cardinality. Notice that S is finite,S = m1, . . . ,mt.Now, we will prove that radM is a proper subset of M by contradiction:assume that radM = M . Then m1 ∈ radM . By the definition of radM ,we have m1 =

∑`j=1 rjnj for some rj ∈ radA and some nj ∈M . If we now

express each nj in the generating set S, we get nj =∑t

i=1 aijmi for some

50

3.6 The radical

aij ∈ A and altogether

m1 =∑j=1

rj

t∑i=1

aijmi =

t∑i=1

∑j=1

rjaij

mi =

t∑i=1

r′imi. (3.1)

Observe that the coefficient r′i =∑`

j=1 rjaij belongs to radA since radA isa right ideal by Proposition 3.18. Now, if t = 1, then we get from equation(3.1) that (1− r′1)m1 = 0. This implies m1 = 0 since 1− r′1 is invertible, acontradiction to the minimality of S. If t > 1, then equation (3.1) can bewritten as (1− r′1)m1 =

∑ti=2 r

′imi, which again contradicts the minimality

of S, since 1− r′1 is invertible. Thus we have shown m1 6∈ radM

Let I be an ideal of A. Then the powers Im are defined inductively Im =IIm−1 =

∑ti=1 aibi | t ≥ 1, ai ∈ I, bi ∈ Im−1. Note that Im is an ideal

of A. The ideal I will be called nilpotent if there exists an integer m ≥ 1such that Im = 0.

Corollary 3.21. The radical radA of a finite-dimensional algebra A is anilpotent ideal.

Proof. Note that we have two possible interpretations of radA, namely theJacobson radical and the radical of A as left A-module. However, it followsdirectly from the definition that they coincide.

Let I = radA. By definition we have Im = rad Im−1, that is, Im equalsthe radical of Im−1, viewed as left A-module. Hence by Nakayama’sLemma 3.20, Im is a proper submodule of Im−1 and therefore a propersubspace. Since A is finite-dimensional, we must have Im = 0 for m ≥dimK A.

We now translate these concepts into the language of categories. We call aK-category C a spectroid if it satisfies the following three properties: (i)the morphism spaces C (x, y) are finite-dimensional, (ii) the endomorphismalgebra C (x, x) is local for each object x and (iii) objects are pairwise non-isomorphic. A spectroid is finite if it has only finitely many objects.

Example 3.22. If A is a finite-dimensional algebra and S a complete set ofpairwise orthogonal, primitive idempotents, then the category AS is a finitespectroid. ♦

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3 Algebras

A morphism f ∈ C (x, y) of a K-category C is called radical if for eachg ∈ C (y, x) the element 1x − gf is invertible in the algebra C (x, x). Theradical morphisms form an ideal which is called the radical of C and willbe denoted by rad C .

Example 3.23. If C is a spectroid then the radical rad C is given by: thesubspace (rad C )(x, x) is the Jacobson radical rad(C (x, x)) of the finite-dimensional algebra C (x, x) for each object x and (rad C )(x, y) = C (x, y)for x 6= y since the objects are pairwise non-isomorphic. It therefore makessense to omit the parenthesis as both ways of interpretation of rad C (x, y)amounts to the same thing. ♦

If I is an ideal of a K-category C its powers are defined by

Im(x, y) =∑z∈C

I (z, y)Im−1(x, z).

Note that Im is again an ideal of C . We say that I is nilpotent ifthere exists some integer m ≥ 1 such that Im is the zero ideal, that is,Im(x, y) = 0 for all x, y ∈ C . The following result will be important in theforthcoming.

Corollary 3.24. If C is a finite spectroid, then rad C is a nilpotent ideal.

Proof. Let x1, . . . , xt be the objects of C . Then for each i = 1, . . . , tthe algebra C (xi, xi) is finite-dimensional and local. Therefore, by Corol-lary 3.21 there exists some mi ≥ 1 such that radmi C (xi, xi) = 0. LetM = maxm1, . . . ,mt. We now argue that rad C ` = 0 for ` ≥ (M + 1)t.Indeed, if f = f` · · · f1 is a composition of ` radical morphisms withfj ∈ C (xhj−1

, xhj ) and ` ≥ (M +1)t, then at least one index j is repeated atleast M + 1 times in the sequence h0, h1, . . . , h`. Hence f can be written inthe form f = vgMgM−1 · · · g1u, where u ∈ C (xh0 , xj), v ∈ C (xj , xh`) and giare endomorphisms gi ∈ C (xj , xj) for i = 1, . . . ,M . Therefore gM · · · g1 = 0and consequently f = 0.

Exercises

3.6.1 Show that the element E(ij) for i 6= j is nilpotent, but is not radical in thealgebra Kn×n. Determine rad(Kn×n).

3.6.2 Calculate the radical of the algebra of lower triangular matrices.

52

3.7 The quiver of an algebra

3.6.3 Show that for a quiver Q, the radical of the category KQ is the ideal gen-erated by the arrows, that is, (radKQ)(x, y) is the set of linear combinations ofpaths from x to y of positive length.

3.6.4 Let C be a K-category with finitely many objects and finite-dimensionalmorphism spaces. Denote by AC the algebra AC =

⊕x,y∈C C (x, y), see Exer-

cise 3.3.5. Show that if I is some ideal of C then I =⊕

x,y∈C I (x, y) is an idealof AC . Prove that I = radAC if I = rad C and more generally I = radmAC ifI = radm C . Also prove that AC/I = AC /I holds in general.


We now set out to define the quiver of an algebra A. As we shall see, therequirement for the field K being algebraically closed will be important. Forthis we first construct a basic algebra B which is Morita equivalent to A.Let S be a complete set of pairwise orthogonal, primitive idempotents ofB and BS the category constructed in Section 3.4. The category BS is afinite spectroid, see Example 3.22. In the following it is explained how toassociate a quiver QC to any spectroid C . The quiver QA of the algebra Ais by definition the quiver QBS . We will have to wait until Remark 4.27 tounderstand why QA does not depend on the choice of S.

Let C be a spectroid. For each pair of objects x, y ∈ C , we choose morphisms

α(x,y)1 , . . . , α(x,y)

nxy ∈ rad C (x, y) (3.2)

which become a K-basis in the vector space rad C (x, y)/ rad2 C (x, y) un-der the canonical projection. Note that by definition C (x, y) is finite-dimensional and therefore each space rad C (x, y)/ rad2 C (x, y) is also finite-dimensional. A morphism f ∈ C (x, y) is called irreducible if it is radicalbut does not belong to the radical square.

The quiver QC associated to the spectroid C has then the objects of C asits vertices and the morphisms (3.2) as arrows from x to y. Note that thecanonical projection Π: KQC → C is always bijective on the objects.

Proposition 3.25. If the field K is algebraically closed and the spectroidC is finite then the canonical projection Π: KQC → C is surjective on themorphism spaces.

Proof. For each object x of C the algebra EndC (x) is local and finite-dimensional. Hence rad EndC (x) = rad C (x, x) is a maximal ideal and

53

3 Algebras

therefore EndC (x)/ rad EndC (x) a skew field extension of K1x which isfinite-dimensional over K. Since K is assumed to be algebraically closedthis implies that each element a ∈ EndC (x) can be written uniquely in theform a = λ1x + r where λ ∈ K and r ∈ rad EndC (x). Observe that thisalready shows that it suffices to prove that each element of the radical rad Cbelongs to the image of Π.

We now prove by induction on ` that for each two objects x, y each ele-ment g of rad` C (x, y) can be written as g = g(`+1) + g(`) where g(`+1) ∈rad`+1 C (x, y) and g(`) is a linear combination of compositions of ` arrowsof the quiver, that is,

g(`) =

N∑i=1

λiρi,`ρi,`−1 . . . ρi,1

for some N , some coefficients λi ∈ K and some arrows ρi,j ∈ (QC )1 fori = 1, . . . , N and j = 1, . . . , `.

The assertion is true for ` = 1 by definition of the arrows (3.2) of QC . Nowassume the assertion to hold for ` − 1 and let g ∈ rad` C (x, y), where x, yare two objects of C . By the definition of the power of an ideal we have

g ∈∑z∈C

rad C (z, y) rad`−1 C (x, z),

that is, there exist elements σz ∈ rad C (z, y) and ηz ∈ rad`−1 C (x, z) suchthat g` =

∑z∈C σzηz. By induction hypothesis, for each z, we may write

ηz = η(`)z + η

(`−1)z and σz = σ

(2)z + σ

(1)z . Hence

g =∑z∈C

σzηz

=∑z∈C

σzη(`)z +

∑z∈C

σ(2)z η(`−1)

z︸︷︷︸=g(`+1)∈rad`+1 C (x,y)

+∑z∈C

σ(1)z η(`−1)

z︸︷︷︸=g(`)

.

This proves the induction argument and hence the result.

Remark 3.26. If K is not algebraically closed, the quiver QC does notencode all information of C , because C (x, x)/ rad C (x, x) might be someproper skew field extension over K and the quiver should be enhanced toa species (to each vertex a division algebra is attached and to each arrowa bimodule over the corresponding skew fields). Dlab and Ringel classified

54


the species of finite representation type in [7] and those which are tame in[8]. ♦

In the situation of Proposition 3.25, we consider I = Ker Π given byI (x, y) = g ∈ KQC (x, y) | Π(g) = 0. It is easy to see that I is anideal of the category KQC .

We call an ideal I of KQC admissible if there exists some m ≥ 2 suchthat radmKQC ⊆ I ⊆ rad2KQC .

Lemma 3.27. Let K be an algebraically closed field, let C be a finite spec-troid with quiver QC and denote by Π: KQC → C the canonical projection.Then Ker Π is an admissible ideal of the category KQC .

Proof. Recall from Exercise 3.6.3 that the radical of KQC is generated bythe arrows. From the construction of the arrows of QC , it follows thatKer Π ⊆ rad2KQC . Also, by construction Π(radKQC ) = rad C and there-fore inductively Π(radmKQC ) = radm C . Since rad C is nilpotent, we con-clude that there exists an integer m ≥ 2 such that radmKQC ⊆ Ker Π.

Putting together what we already know, we obtain the following result.

Theorem 3.28 (Gabriel). Let A be a finite-dimensional algebra over analgebraically closed field K. Then A is Morita equivalent to an algebra AQ/I,where I is an admissible ideal in the path algebra AQ of a finite quiver Q.

Proof. By Proposition 3.16, there exists a basic algebra B such that A andB are Morita equivalent. Since B is basic, there exists a complete set S ofpairwise orthogonal, primitive idempotents, such that BS is a finite spec-troid. Let Q be the quiver of the spectroid BS and I = Ker Π, whereΠ: KQ → BS is the canonical projection. By Proposition 3.25, Π is sur-jective on morphisms and by Lemma 3.27 the ideal I is admissible in KQ.Hence Π induces a functor KQ/I → BS which is bijective on objects andon morphisms, hence an isomorphism of categories.

Consequently we have the following equivalence of categories modKQ/I 'modBS ' modB ' modA, where the second equivalence is due to Proposi-tion 3.11 and the last to Morita equivalence. Finally, if we denote by AQ =KQ the path algebra and by I =

⊕x,y∈Q I (x, y) ⊂ AQ the ideal which

corresponds to I , then I is admissible in AQ and modAQ/I ' modKQ/Iby Exercise 3.3.5 and 3.6.4.

55

3 Algebras

Remark 3.29. Theorem 3.28 explains the expression “let A = KQ/I bean algebra” which is quite common in the literature. ♦

Exercises

3.7.1 Let Q be a finite quiver. Show that every admissible ideal I of AQ is finitelygenerated.

3.8 Relations

Since admissible ideals are finitely generated, see Exercise 3.7.1, it is enoughto give a finite set of generators ρ1, . . . , ρt which are usually written aselements of the path category, that is, each ρi is a linear combination ofpaths with the same starting vertex and the same terminating vertex. Suchgenerators are called relations. Relations which are a single path are calledzero relation. A quiver together with an admissible ideal (or a set ofrelations defining it) is called bounded quiver.

Remark 3.30. By Theorem 3.28 we know how to associate a quiver Q to analgebra A. As we shall see in Chapter 4 the quiver is uniquely determined byA. However, the relations are not uniquely determined by A. For example,let Q be the quiver as shown in the next picture.

rrrr

@@R

@@R

α

β

γ

δ

Clearly, the two sets of relations R1 = γα, δβ and R2 = γα+ δβ, δβ aredifferent, but they generate the same ideal. But it can even happen thatdifferent ideals generate isomorphic algebras, for instance if I1 = 〈γα〉 andI2 = 〈δβ〉. Then the two algebras KQ/I1 and KQ/I2 are isomorphic. ♦

Theorem 3.28 shows that to study modules over finite-dimensional algebrasit is enough to study modules of quotients of path algebras. We can deriveeven more: it is enough to study quiver representations, as shows the nextresult. For this, we need some more terminology.

A covariant, K-linear functor F : KQ → vec is said to satisfy the ideal Iof KQ if F (ϕ) = 0 for each ϕ ∈ I . If V is a representation of Q, then wesay that V satisfies the ideal I if Vρ = 0 for each ρ ∈ I .

56

3.8 Relations

If D and C are categories, we call D a subcategory of C if the objects ofD form a subclass of the objects of C and for each two objects x, y of D , theset D(x, y) is a subset of C (x, y) and the composition of D is induced by thecomposition in C , that is, if f ∈ D(x, y) and g ∈ D(y, z) then gD f = gC f ,where D (resp. C ) denote the composition in D (resp. in C ). We writeD ⊆ C to indicate that D is a subcategory of C .

A subcategory D ⊆ C is called full if D(x, y) = C (x, y) for any two objectsx, y of D . Note that in order to define a full subcategory D of C it is enoughto specify the class of objects of D .

Proposition 3.31. The category modKQ/I is isomorphic to the full sub-category repI Q of repQ given by the representations which satisfy the idealI .

Proof. The canonical projection functor p : KQ → KQ/I defines a func-tor p∗ : mod(KQ/I ) → mod(KQ), F 7→ F p. Note that p∗ induces anisomorphism between modKQ/I and the full subcategory modI (KQ) ofmod(KQ) of functors which satisfy the ideal I .

We know from Exercise 2.4.1 that mod(KQ) is isomorphic to repQ. Hencethe full subcategory modI (KQ) corresponds to a full subcategory of repQ,which clearly is repI Q.

Example 3.32. Consider the example, where B = A/I is the quotient ofthe algebra of lower triangular matrices of size 5×5 and I the ideal generatedby E(i,j) for i−j > 1. We have then I = rad2A, where radA is the Jacobsonradical of A. The quiver Q is the same for CB as for CA, namely linearlyoriented with 5 vertices, and the admissible ideal is generated by the pathsαiαi−1 for i = 2, 3, 4, indicated in the picture with dotted arcs.

Q : r r r r r- - - -1 2 3 4 5α1 α2 α3 α4

Thus in repI Q, we have only the following indecomposable representationsup to isomorphism (with irreducible morphisms between them):

@@R @@R @@R @@R[5, 5]

[4, 5]

[4, 4]

[3, 4]

[3, 3]

[2, 3]

[2, 2]

[1, 2]

[1, 1]

(3.3)

because the others listed in (2.6) do not satisfy all the relations. ♦

57

3 Algebras

Proposition 3.31 and the previous example could give the idea that themethod of describing the modules over an algebra is a straightforward pro-cess in three steps: first one determines the quiver Q of A and the admissibleideal I such that KQ/I is Morita equivalent to A, second one calculates allindecomposable representations of Q and in the third step one eliminatesthose representations which do not satisfy I. Indeed, theoretically this iscorrect. However, in practice it is not always easy to calculate Q and inmost examples it will be completely impossible to achieve the second step:there are only few quivers for which this is possible, the vast majority arewild, that is, there are two-parameter families of pairwise non-isomorphicindecomposable representations for most quivers.

Exercises

3.8.1 How many isomorphism classes of indecomposable left A-modules exist if

A = K−→A 6/I, where I = rad4K

−→A 6?

3.8.2 Define what formally is a “two-parameter family of indecomposables”.

3.9 Summary

We worked quite hard to get Theorem 3.28, which is the main result ofthis chapter. We even worked harder than really necessary, because we de-veloped ways back and forth between algebras and categories. Althoughthis might seem unnecessary for now, it will pay off later, since we canapply the concepts in much broader context. In particular, we can speakof the radical of any K-category and know that the concept generalizes theJacobson radical of an algebra.

In the special situation that K is an algebraically closed field and A a ba-sic finite-dimensional K-algebra, Proposition 3.31 summarises that each A-module M ∈ modA can be viewed as a representation of the quiver of A.You will have to wait until the end of the next chapter, see Remark 4.27,to see that this quiver is uniquely determined by A up to isomorphismsof quivers, where an isomorphism of quivers ϕ : Q → Q′ is a pair ofbijections ϕi : Qi → Q′i for i = 0, 1 such that sQ′(ϕ1α) = ϕ0(sQα) andtQ′(ϕ1α) = ϕ0(tQα) for all α ∈ Q1.

We have exposed three different languages for the same subject, and sum-marise them for the case, when K is algebraically closed.

58

3.9 Summary

In the classical point of view we have algebras (to be thought as finite-dimensional over the field K) together with modules (again finite-dimensio-nal). In the categorical point of view we consider finite spectroids togetherwith functors (covariant and K-linear into the category of finite-dimensionalvector spaces over K). In the combinatorial point of view, we consider abounded quiver (a quiver together with an admissible ideal) and repre-sentations (of that quiver satisfying the given admissible ideal).

Thus, if A is such an algebra, C = CA the corresponding spectroid withquiver Q and admissible ideal I, we have that

modA ' mod C ' repI Q

are three equivalent categories. It is common to speak of the quiver of analgebra or to look at “full subcategories of the algebra A”, which just meansthat one is adopting the categorical point of view, where others would preferto look at eAe, for e some idempotent, meaning exactly the same thing. Inthe forthcoming, we will freely move between these three languages.

If you compare different articles on representation theory you will find thatnone of what we have said is really standard; it begins with the fact thatsome authors favour the notation vw for the composition of two paths vfrom i to j and w from j to h (we denoted it as wv just as functions),some authors even go so far as to denote the composition of functions andhomomorphisms in that way, although most of them denote functors andmorphisms of categories as we do. It is also a matter of taste whether rightor left modules are favoured, whereas most authors prefer covariant functorsover contravariant ones, representations over corepresentations and equiv-alences over antiequivalences. The word “spectroid” is almost completelyabsent in the literature, but we will use it since it is useful.

As you can see and could have expected, representation theorists have ex-panded into this vast field of possibilities turning it into a political landscape,where a consensus clearly is not at hand. In some articles definitions are leftsufficiently unspecific in order to turn it into a nice exercise for the readerto work his or her way through it to see whether a consistent interpretationis possible.

59

3 Algebras

60

Chapter 4

Module categories

The focus is now changed towards modules. Several categorical notions formodules are developed. In each case a full characterization of indecompos-able modules with that particular property is achieved. This is a first steptowards understanding the structure of module categories.

4.1 The categorical point of view

The problem of determining the modules over a given algebra lies in the coreof representation theory. It is though often useful to consider all modulesor all modules up to isomorphisms together with all morphisms betweenthem. The morphisms give structural insight, they are the flesh of themodule category. Without them, the modules would just be a bare list ofdisconnected entities.

Recall that for a given finite-dimensional algebra A, we denote by modAthe category of all finite-dimensional left A-modules. We start with someeasy remarks on modA.

In modA the surjective and injective homomorphisms satisfy the followingcancellation properties, whose proof is straightforward from the definitionof surjectivity and injectivity.

Lemma 4.1 (Cancellation properties). Let f : M → N be a homomorphismof left A-modules. Then f is surjective if and only if for any two homomor-phisms g1, g2 : N → P we have that g1f = g2f implies g1 = g2. Similarly,f is injective if and only if for any two homomorphisms g1, g2 : L → M wehave that fg1 = fg2 implies g1 = g2.

61

4 Module categories

The cancellation conditions stated in Lemma 4.1 do not involve elements ofthe module M or N . Instead they use the relationship of f with respectto all other morphisms. The cancellation properties can be stated in everycategory. A morphism f ∈ C (x, y) for which g1f = g2f always implies g1 =g2 is called an epimorphism (or just epi for short), whereas if fg1 = fg2

always implies g1 = g2 then f is called a monomorphism (or mono). Thus,“injectivity” is an element-based property whereas mono is a categoricalproperty. We shall meet many categorical properties in this chapter and itwill be useful to see what they mean in different categories.

Remark 4.2. In the categorical point of view all properties have to be ex-pressed using morphisms only. While it is instructive to rephrase propertiesin categorical language, since it broadens the view to greater generality, italso can conduct to excess in abstraction loosing the grip of concrete exam-ples. Caution is thus required to achieve a reasonable balance between bothviews. ♦

Exercises

4.1.1 Let Q be a finite quiver and f : V →W a morphism of representations of Q.Show that f is mono if and only if fx is injective for each vertex x of Q. Similarlyshow that f is epi if and only if fx is surjective for all x.

4.1.2 Let A be a finite-dimensional K-algebra. Show that a morphism f : M → Nis epi (resp. mono) in modA if and only if it is surjective (resp. injective). Usethat the image B = f(M) is a submodule and that the quotient N/B is again anA-module. Then compare the canonical projection π : N → N/B with the zeromap 0: N → N/B.

4.1.3 Prove that the inclusion Z→ Q is an epimorphism in the category of rings.This shows, that even in concrete categories, where the objects are sets withsome structure and morphisms are functions, which preserve that structure, epi-morphisms must be surjective.

4.2 Duality

We first look at another categorical notion: an object x ∈ C is called initialobject if for each object y ∈ C the morphism set C (x, y) consists of exactlyone element. Similarly, x is called terminal object if for each y ∈ C theset C (y, x) has exactly one element.

62

4.2 Duality

As you can see, categorical notions come in pairs which correspond to eachother: monomorphisms correspond to epimorphims and initial objects toterminal objects. If you look at the definitions, you should see that onecan be obtained from the other by somehow reversing the direction of thearrows. In the following we will formalize this.

The opposite category C op of C is defined as follows: it has the sameobjects as C , its morphism sets are C op(x, y) = C (y, x) and the compositionis given by

C op(y, z)× C op(x, y)→ C op(x, z), (g, f) 7→ g op f = f g.

We then see, that a morphism f : x→ y is mono in C if and only if f : y → xis epi in C op. Similarly an object x is initial in C if and only it is terminalin C op.

Defining the opposite category has economical advantages, since we onlyhave to define half of the concepts: instead of stating the definition of ter-minal objects we could have said that terminal is the dual concept toinitial.

However, it is not straightforward how we could dualize properties in aconcrete category like modA, as it is not clear that (modA)op is again amodule category. The following result solves this problem. The oppositealgebra Aop is A as abelian group, the multiplication a ·op b = ba, wherethe latter denotes the usual product in A. Furthermore, Aop has the sameinclusion of K in the center Z(Aop) = Z(A). Furthermore, for a vector spaceV we denote by DV = HomK(V,K) the dual space.

Proposition 4.3. Let A be a finite-dimensional K-algebra. Then the dual-ization yields an isomorphism of categories D: (modA)op → mod(Aop).

Proof. For each left A-module M , define Φ(M) = DM = HomK(M,K), thedual space of M . Now define the multiplication

Aop ×DM → DM, (a, ϕ) 7→ aϕ

by aϕ : M → K,m 7→ ϕ(am). Clearly this multiplication satisfies the distri-bution laws (a+ b)ϕ = aϕ+ bϕ and a(ϕ+ ψ) = aϕ+ aψ. Also

(a(bϕ))(m) = (bϕ)(am) = ϕ(bam) = (ba)ϕ(m) = (a ·op b)ϕ(m).

Note that for each finite-dimensional vector space M the double dual DDMis canonically isomorphic to M , via the map M → DDM , m 7→ (ϕ 7→ ϕ(m)).Hence we see that Φ has an inverse which is also given as dualization.

63

4 Module categories

Remarks 4.4 (a) Since for infinite-dimensional vector spaces V the dou-ble dual DDV is not isomorphic to V , the category (ModA)op, that is, theopposite of the category of all A-modules, cannot be expected to be isomor-phic to ModAop. In fact, one can show that (ModA)op is never a modulecategory.

(b) If we view A as a category AS for S = e1, . . . , et some complete setof pairwise orthogonal, primitive idempotents, then we can view Aop simplyas (AS)op which is the same as (Aop)S . Therefore, if Q is the quiver of A,then Aop has the opposite quiver Qop of Q — it has the same vertices asQ but the direction of the arrows is reversed, that is, the two functions sand t of Q are interchanged.

An Aop-module M ∈ mod(Aop), if viewed as representation of Qop, is there-fore mapped to the representation DM of Q which has the space D(Mi)assigned to the vertex i and if α : i→ j is an arrow in Q, then αop : j → i isan arrow of Qop and

(DM)α = D(Mαop) : DMi → DMj , ϕ 7→ ϕ Mαop

describe the linear maps of the representation associated to the arrows ofQ. ♦

Exercises

4.2.1 Determine the objects in the category of sets which are initial and also thosewhich are terminal.

4.2.2 Show that the zero module, which is the zero vector space, is both initialand terminal in the category modA.

4.3 Kernels and cokernels

An object z of a category C is called zero object if for each x ∈ C thesets C (x, z) and C (z, x) contain precisely one element. If there exists azero object z in C , then for any pair x, y of objects in C there exists adistinguished element, called the zero morphism and denoted by 0, whichis obtained by composing the unique element of C (x, z) with that of C (z, y).

Example 4.5. For each algebra A the module category contains a zeroobject: the space 0, called the zero module, denoted by 0. ♦

64


Let C be a category with a zero object. Let f ∈ C (x, y). Then a kernel off is a map g ∈ C (t, x) such that fg = 0 satisfying the universal propertythat, if g′ ∈ C (t′, x) with fg′ = 0, then there exists a unique h ∈ C (t′, t)such that gh = g′. This is usually expressed as a diagram as shown in thefollowing picture.

t

t′

x y

- -

6

g f

g′∃!h

The existence of kernels is not generally guaranteed. There are categorieswhere kernels always exist and others where they do not, see Exercise 4.3.1.Let f : x→ z be a morphism in a category C and y an object of C . We saythat f factors over y if there exist morphisms g : x→ y and h : y → z suchthat f = hg. If g : x→ y (or h : y → z) is given, then we say that f factorsthrough g (resp. factors through h) and call h (resp. g) a factorizationof f .

Example 4.6. In the module category modA each morphism f : M → Nadmits the inclusion ι : f−1(0) → M as kernel: First, note that the setf−1(0) = m ∈ M | f(m) = 0 is a submodule of M . Clearly fι = 0 holdsand if g′ : L′ → M satisfies fg′ = 0 then g′(L′) ⊆ f−1(0) yields the desiredfactorization through ι which is unique since ι is injective. ♦

The dual concept of kernel is cokernel. Using the existence of kernels inmodA, that is, the existence of a kernel for each morphism, we get fromProposition 4.3 that there exists also a cokernel for each morphism. Al-though this might seem satisfactory from a categorical point of view, weshall describe the construction of cokernels in modA explicitely.

Lemma 4.7. For each morphism f : M → N in modA the canonical pro-jection π : N → N/ Im f is a cokernel of f , where Im f = f(m) | m ∈ Mis the image of f .

Proof. Clearly πf = 0. To verify the universal property, let h : N → Pbe a morphism with hf = 0. Then we have h(n) = 0 for each n ∈ Im f .Therefore, the K-linear map h′ : N/ Im f → P , h′(n + Im f) = h(n) is welldefined. This map is also clearly an A-module homomorphism and satisfiesh′π = h. That h′ is unique with that property follows from the fact that πbeing surjective is epi.

65

4 Module categories

Lemma 4.8. Let C be a category with zero object and f ∈ C (x, y) a mor-phism. If g1 ∈ C (t1, x) and g2 ∈ C (t2, x) are two kernels of f then t1 andt2 are isomorphic.

Proof. By the universal property of g1 there exists a unique h1 ∈ C (t2, t1)such that g1h1 = g2. Similarly, there exists a unique h2 ∈ C (t1, t2) such thatg2h2 = g1. Now h1h2 ∈ C (t1, t1) satisfies g1(h1h2) = g1 = g11t1 . Again bythe universal property of g1 we must have h1h2 = 1t1 . Similarly h2h1 = 1t2 .This shows that t1 and t2 are isomorphic.

Because of this result, often the object t is called kernel instead of the mor-phism g ∈ C (t, x). Dualizing we obtain that two cokernels are isomorphic.We have seen that in modA each morphism has a kernel and a cokernel.It is common to use the notation Ker f (resp. Coker f) for a chosen kernel(resp. cokernel) of a morphism f .

Lemma 4.9. Let f : M → N be a morphisms in modA. Then f is injectiveif and only if Ker f = 0 and dually, f is surjective if and only if Coker f = 0.

Proof. If f is injective, then each fiber consists of at most one element. Sincef(0) = 0 always holds, Ker f = f−1(0) = 0 = 0 follows. Conversely, iff−1(0) = 0, then f(m) = f(m′) implies m−m′ ∈ f−1(0). This shows thatf is injective.

We know from Lemma 4.1 that f is injective if and only if f is a monomor-phism. Hence f is a monomorphism if and only if Ker f = 0. Since theseare categorical properties we can “dualize” them and get the result: themorphism Df : DN → DM in the category modAop is a monomorphism ifand only if Ker(Df) = 0. But Df is a monomorphism if and only if f is anepimorphism, which we know is equivalent to f being surjective. Similarly,Ker(Df) = 0 holds if and only if Coker f = 0.

For f : M → N we choose a fixed kernel and denote it by Ker f . Similarly,we choose a cokernel and denote it by Coker f .

We now can iterate or combine these two constructions. If g : U → M is akernel of f : M → N and h : V → U is a kernel of g then we clearly havef(gh) = 0. It follows therefore from the universal property of g that h = 0.But then V ' 0 follows, since h0 = h1V implies by the universal property ofh that 0 = 1V . This shows that the kernel of a kernel is always zero. Duallythe cokernel of a cokernel is zero. Concerning the kernel of a cokernel andthe cokernel of a kernel we prove the following result.

66


Lemma 4.10. Let C be a category in which each morphism has a kernel anda cokernel. Then, for each morphism f : M → N there exists a canonicalmorphism fι : Coker(Ker f)→ Ker(Coker f).

Proof. Denote by g : L → M a kernel of f and by h : N → P a cokernel off . Now we use the universal property of the cokernel Coker g : M → M ′:since fg = 0 there exists a unique ξ : M ′ → N such that ξ Coker g = f ,see the following picture for illustration.

M N

M ′ N ′

L P- - -

?

6

-

g f h

Coker g Kerhξ

fι

Since Coker g is an epimorphism, we get from hξCoker g = hf = 0 thathξ = 0. Using now the universal property of Kerh : N ′ → N , we obtain aunique morphism fι : M

′ → N ′ such that Cokerh fι = ξ.

The morphism (or the object) Ker(Coker f) is called image of f and denotedby Im f . In the category modA the image Im f of a morphism f : M → Nis isomorphic to the inclusion f(M) → N . Similarly Coker(Ker f) is calledcoimage and denoted Coim f .

Lemma 4.11. If C = modA then for each morphism f the canonical mor-phism of Lemma 4.10 is the isomorphism fι : M/f−1(0)→ Im f .

Proof. We then know that Ker f is isomorphic to the inclusion of f−1(0) inM and consequently Coker(Ker f) : M → M/f−1(0) is the canonical pro-jection. We also know that Coker f is isomorphic to the canonical pro-jection N → N/ Im f . Therefore Ker(Coker f) is isomorphic to the in-clusion Im f → N . By construction, fι : M/f−1(0) → Im f is given byfι(m + f−1(0)) = f(m). Clearly fι is injective and surjective. Hence theresult follows.

A category C is additive if the morphism sets are abelian groups suchthat the composition is bilinear, there exists a zero object and for any twoobjects in C there exists a direct sum. In an additive category, kernels aremono: if g : U → M is a kernel of f : M → N then for any two morphismsh, h′ : W → U with gh = gh′ we have g(h − h′) = 0 and thus by the

67

4 Module categories

uniqueness of the factorization h− h′ = 0, showing that g is mono. Duallycokernels are epi in an additive category.

A category C is abelian if it is additive and for any morphism exists akernel, a cokernel and the canonical map from a coimage to an image is anisomorphism. We have seen that the category modA is abelian.

Given two morphisms with the same codomain, say fX : X → Z andfY : Y → Z. Then a pull-back of fX and fY is a triple (U, gX , gY ), whereU is an A-module and gX : U → X, gY : U → Y are morphisms such thatfXgX = fY gY , with the universal property, that for any other such triple(V, hX , hY ) there exists a unique morphism ζ : V → U such that hX = gXζand hY = gY ζ. The following diagram visualizes the situation.

Y

X

Z

U

V

? ?-

-

HHHHH

HHHHj

AAAAAAAAAU

R

fX

fY

gY

gX

∃!ζhX

hY

Note that pull-backs exist in modA, see Exercise 4.3.4. Furthermore, theobject U is unique up to isomorphism, see Exercise 4.3.5 and will be de-noted by X

∏f,g Y . The dual concept of pull-back is called push-out, see

Exercise 4.3.6.

Exercises

4.3.1 Let V be the full subcategory of vec given by the vector spaces of evendimension. Show that there exists a morphism f : K2 → K2 in V such that akernel exists in V . Give another morphism where no kernel exists.

4.3.2 Let Q be a finite quiver and f : V → W a morphism of representationsof Q. Define the representation L of Q by Lx = f−1

x (0) for each vertex x andLα : Lx → Ly to be the maps induced by Vα. Show that the inclusion L→M is akernel of f . Construct the image and the cokernel of f explicitely.

4.3.3 Let C be an abelian category and M,N be two objects such that thereexist two morphisms π : M → N and ι : N → M satisfying πι = idN . Show thatM = N ⊕N ′, where ι′ : N ′ →M is a kernel of π.

68

4.4 Unique decomposition

4.3.4 Let C be an abelian category and fX : X → Z, fY : Y → Z be two mor-phisms. Denote [fX −fY ] : X⊕Y → Z the obvious map. Show that Ker[fX −fY ]yields a pull-back of fX and fY .

4.3.5 Show that if (U, gX , gY ) and (U ′, g′X , g′Y ) are two pull-backs of fX : X → Z,

fY : Y → Z, then there exists a unique isomorphism ϕ : U → U ′ such that g′Xϕ =gX and g′Y ϕ = gY . You may use the uniqueness of kernels and Exercise 4.3.4 orgive a direct argument for this.

4.3.6 State the definition of a push-out (including the universal property) andprove existence and uniqueness using dualization.

4.3.7 Let g : Z ′ → Z and b : Y → Z be two morphisms in an abelian category C .Furthermore, let a : X → Y be a kernel of b. Show that there exists some morphismξ : Z ′ → Y with g = bξ if and only if (X ⊕Z ′, [0 1], [a ξ]) is a pull-back of g and b.


As usual A denotes a finite-dimensional K-algebra. Any M ∈ modAadmits a decomposition into a direct sum of indecomposable modulesM =

⊕si=1Mi, since either M is already indecomposable, or it decom-

poses, say M ' N1 ⊕N2, and then by induction on the dimension we knowthat N1 and N2 admit such a decomposition.

Such a direct sum decomposition M =⊕s

i=1Mi comes equipped withcanonical inclusions ιi : Mi → M and projections πi : M → Mi such thatπiιi = 1Mi , πiιi′ = 0 for i 6= i′ and

∑si=1 ιiπi = 1M . Let M ′ =

⊕tj=1M

′j with

canonical inclusions ι′j and projections π′j . Then each morphism ϕ : M →M ′

gives rise to a family of morphisms ϕji = π′jϕιi : Mi → M ′j . Note that wecan recover ϕ from this family since ϕ =

∑j,i ι′jϕjiπi.

Now, we want to see how these families behave under the composition ofmorphisms. For this, let M ′′ =

⊕uk=1M

′′k be yet another module with the

corresponding canonical inclusions ι′′k and projections π′′k and ψ : M ′ →M ′′

69

4 Module categories

a morphism and ψkj = π′′kψι′j : M ′j →M ′′k . Then

ψϕ =

∑k,j′

ι′′kψkj′π′j′

∑j,i

ι′jϕjiπi

=∑k,i

ι′′k

∑j′,j

ψkj′π′j′ι′jϕji

πi

=∑k,i

ι′′k

∑j

ψkjϕji

πi,

which shows that (ψϕ)ki =∑

j ψkjϕji, exactly like matrix multiplication.

Therefore, we can write morphisms between direct sums as matrices betweenthe indecomposables of the given decomposition. The following result showsthat the decomposition into indecomposables is unique up to order andisomorphisms.

Theorem 4.12 (Krull-Remak-Schmidt). If M =⊕s

i=1Mi is isomorphicto N =

⊕tj=1Nj, where all Mi and Nj are indecomposable, then t = s and

there exists a permutation σ of 1, . . . , t such that Mi is isomorphic to Nσ(i)

for i = 1, . . . t.

Proof. Let ϕ : M → N be an isomorphism. Then ϕ can be written as matrixϕ = (ϕji)

tj=1

si=1, where ϕji : Mi → Nj . Similarly write ψ = ϕ−1 as matrix

and observe that idM1 = (ψϕ)11 =∑t

l=1 ψ1lϕl1. One summand on the rightmust then be invertible, since EndA(M1) is local. We may assume withoutloss of generality that ψ11ϕ11 is invertible (otherwise reorder N1, . . . , Nt).But, since both modules, M1 and N1, are indecomposable, we have thatboth homomorphisms, ψ11 and ϕ11, are invertible.

We now exchange the given isomorphism ϕ by ϕ′ = αϕβ, where α and βare defined as

α =

idN1 0 · · · 0

−ϕ21ϕ−111 idN2 · · · 0

......

· · ·...

−ϕt1ϕ−111 0 · · · idNt

70


and

β =

idM1 −ϕ−1

11 ϕ12 · · · −ϕ−111 ϕ1s

0 idM2 · · · 0...

...· · ·

...0 0 · · · idMs

,and observe that ϕ′ =

[ϕ11 00 Φ

], where Φ:

⊕si=2Mi →

⊕tj=2Nj . Since ϕ′

is bijective, so must be Φ, which is thus an isomorphism again. The resultfollows now by induction.

Remark 4.13. Sometimes it is useful to consider just the indecompos-able modules together with all irreducible morphisms between them. In thepresent section we have found that, from a certain perspective, this is enoughto know all of the module category. Exercise 4.4.1 gives more evidence forthis. That this perspective is too simple shows the following example: con-sider the quiver Q with a single vertex and no loop, then, from that pointof view, the study of repQ ' modK ' vec which is just the theory of linearalgebras of finite-dimensional vector spaces, would be reduced to the studyof the field K together with all its endomorphisms, which are given by scalarmultiplication. ♦

Still in view of the reduction of the module category “up to isomorphism” thefollowing considerations are interesting. We say that a category C is smallif its objects form a set. A skeleton of a category C is a full subcategoryof C whose objects consist of representatives of the isomorphism classesof the objects. In other words if we choose for each isomomorphism class arepresentative and form the full subcategory of C with these representatives,then we get a skeleton of C . Note that two skeletons of C are isomorphic,see Exercsie 4.4.4.

Lemma 4.14. For each finite-dimensional K-algebra A the category modAhas a small skeleton.

Proof. Every A-module M is a K-vector space. Thus there exists a K-linear bijection ϕ : M → Kd, where d = dimKM . Moreover Kd becomes anA-module isomorphic to M if we define the external multiplication

A×Kd → Kd, (a,m) 7→ ϕ(aϕ−1(m)).

This shows that each A-module can be represented, up to isomorphism, bya vector space Kd together with a multiplication A×Kd → Kd. Hence theisomorphism classes of modA form a set.

71

4 Module categories

Exercises

4.4.1 Show that a homomorphism ϕ :⊕s

i=1Mi →⊕t

j=1Nj is radical if and onlyif for each i and j the homomorphism ϕji : Mi → Nj is radical.

4.4.2 Determine a basis for Hom(V,W ), where V = [3, 4]⊕ [1, 3] and W = [2, 4]⊕[4, 5] are A−→A 5

-modules in the notation established in in Section 2.6.

4.4.3 Let A = K−→A 5 and let M =

⊕5i=1[i, 5], where [i, j] denotes an indecompos-

able A-module in the notation established in Section 2.6. Show that the endo-morphism algebra EndA(M) is isomorphic to the algebra B of upper triangularmatrices of size 5× 5. Show also that the linear map B → A,M → M> yields anisomorphism of algebras B → Aop.

4.4.4 Prove that two skeletons of a fixed category C are isomorphic categories.

4.5 Projectives and injectives

We now consider a new categorical concept with its dual counterpart. Anobject P of a category C is called projective, if for any epimorphismf : M → N and any morphism g : P → N , there exists a (usually notunique) morphism h : P →M such that g = fh. It is common to show thisproperty in a diagram as follows.

M N

P

?-

∀g

∀f (epi)

∃h

In particular, we are interested in the objects of modA which are projective,where A is some finite-dimensional K-algebra. Such objects are called pro-jective A-modules. The dual concept of projective is injective, see alsoExercise 4.5.2.

Example 4.15. Let A be a finite-dimensional K-algebra. Then A con-sidered as left A-module is projective in modA. Indeed, if f : M → Nis an epimorphism in modA, then f is surjective by Exercise 4.1.2. Notealso, that any morphism h : A → M is completely determined by the el-ement m = h(1A) ∈ M , since h(a) = ah(1A) = am. We must findm ∈ M such that f(m) = g(1A). For this, let n = g(1A) and ob-serve that, since f is surjective, the fiber f−1(n) is not empty. Choose

72


m ∈ f−1(n) and set h(a) = am. Then h is a homomorphism of A-modulesand fh(a) = f(am) = af(m) = an = ag(1A) = g(a). ♦

Lemma 4.16. Let C be a K-category and P an object which is the directsum P = P1 ⊕ P2. Then P is projective if and only if P1 and P2 areprojective.

Proof. As usual, for i = 1, 2 denote by πi : P → Pi the canonical projec-tion and by ιi : Pi → P the canonical inclusion. Let f : M → N be anepimorphism in C .

Suppose that P is projective. Then for any g : Pi → N we get a morphismgπi : P → N , which by hypothesis factors through f , say gπi = fh for someh : P → M . Now, consider hιi : Pi → M . By construction, g = gπiιi =f(hιi) is the desired factorization of g. This shows that Pi is projective.

Now assume that P1 and P2 are both projective and let g : P → N be anymorphism. Then, by hypothesis the morphism gι1 : P1 → N factors throughf , say gι1 = fh1 for some h1 : P1 → M . Similarly gι2 = fh2 for someh2 : P2 → M . Now define h : P → M by h = h1π1 + h2π2. Then fh =fh1π1 + fh2π2 = gι1π1 + gι2π2 = g, which shows that P is projective.

Proposition 4.17. Each indecomposable projective A-module is isomorphicto a direct summand of A.

Proof. Let P be an indecomposable A-module which is projective. Clearlythere exists a surjective morphism f : An → P for n = dimK P : take abasis p1, . . . , pn of P considered as K-vector space and then define f byf(a) =

∑i aipi for all a = (a1, . . . , an) ∈ An.

Since P is projective there exists a factorization of the identity mapidP : P → P , say idP = fh for some h : P → An. By Exercise 4.3.3, Pis a direct summand of An. Hence An = P ⊕ Q and if Q is decomposedinto indecomposables, we get a decompositon of An into indecomposables.If A =

⊕ti=1 Pi is a decomposition into indecomposables we also get that

An =⊕t

i=1 Pni is a decomposition into indecomposables. Hence, by the

uniqueness of the decomposition, Theorem 4.12, we get that P must beisomorphic to some Pi.

Corollary 4.18. Let A be a basic, finite-dimensional K-algebra. Ife1, . . . , et is a complete set of pairwise orthogonal, primitive idempotentsof A, then Ae1, . . . , Aet is a complete list of pairwise non-isomorphic inde-composable projective A-modules up to isomorphism.

73

4 Module categories

Proof. Clearly, the left A-module A can be decomposed as A =⊕t

i=1Aei.By Example 4.15 and Lemma 4.16 each direct summand Aei is projective.

We first show that Aei is indecomposable: Each f ∈ EndA(Aei) is deter-mined by the value x = f(ei) ∈ Aei and x = f(ei) = f(e2

i ) = eif(ei) = eix ∈eiAei. Now, if x is not invertible in eiAei then it is nilpotent, since by Corol-lary 3.15 the algebra eiAei is local. Therefore xn = 0 for some n and thusfn(ei) = xn = 0 implies fn = 0 and hence f is nilpotent. If, on the otherhand, x ∈ eiAei is invertible, say xy = yx = ei, then g : Aei → Aei, aei 7→ aydefines an inverse of f . Hence EndA(Aei) is local and therefore Aei inde-composable, by Proposition 3.13.

By Proposition 4.17, each indecomposable projective A-module is isomor-phic to some direct summand of A and since A =

⊕ti=1Aei is a decompo-

sition into indecomposables, it follows from the uniqueness of the decom-position, Theorem 4.12, that each indecomposable projective A-module isisomorphic to Aei for some i. It remains to see is that Ae1, . . . , Aet arepairwise non-isomorphic: to do so, assume otherwise. Then there exist iso-morphisms f : Aei → Aej and g : Aej → Aei which are inverse to eachother. Let x = f(ei) ∈ Aej and y = g(ej) ∈ Aei. Then x = xej andtherefore ei = gf(ei) = g(x) = xg(ej) = xy and similarly yx = ej . Thenx : ej → ei and y : ei → ej are inverse isomorphisms in the category AS ,where S = e1, . . . , et, thus ei and ej are isomorphic objects in AS andhence A is not basic, in contradiction to our assumption.

Example 4.19. Let A be the matrix of lower triangular matrices in Kn×n.Then E(11), . . . ,E(nn) is a complete set of pairwise orthogonal, primitiveidempotents. Therefore AE(11), . . . , AE(nn) is a complete list of pairwisenon-isomorphic indecomposable projective A-modules. Note that AE(ii) inthe language of representations of Section 2.6 is denoted [i, n]. ♦

Remarks 4.20 (a) We now translate the description of the indecompos-able projective A-modules into the other languages: recall that for the com-plete set S = e1, . . . , et of pairwise orthogonal, primitive idempotents wehave associated a spectroid AS . Now, the module Aei corresponds to thefunctor AS(ei,−) : AS → vec in the categorical language and to the represen-tation which associates to the vertex ej the vector space AS(ei, ej) = ejAei.

(b) If the algebra A is given by a bounded quiver Q and relations, thatis, A = KQ/I, then to each vertex x, there corresponds a projective in-decomposable representation, which is denoted by Px. To the vertex y the

74


representation Px associates the vector space (Px)y = ey(KQ/I)ex and toan arrow α : y → z the linear map ey(KQ/I)ex → ez(KQ/I)ex, γ 7→ αγ. ♦

Example 4.21. Let A = KQ/I, where Q is the following quiver:

r rr

@@@R

α

1

β γ

δ3 2

with the relations α2 = 0 and γα = 0. For the projective P1, we obtain thevector spaces (P1)1 = Kid1⊕Kα, (P1)2 = Kγ and (P1)3 = Kβ⊕Kβα⊕Kδγand hence for P1 up to isomorphism the representation as shown below onthe left.

K3 K

K2

@@@R

[0 01 0

][1 00 1

] [1 0

]0

01

1

1 2 3

33

@@

γ

δ

βα

β

It is common and usually more efficient to write down a diagram, as shownabove on the right: the vertices are the elements of a basis as representation.In our example these would be 1, α, β, βα, δγ, γ. But instead of writing thesebasis vectors only the corresponding vertex of the quiver is denoted: 1 for1, α ∈ (P1)1, then 2 for γ ∈ (P1)2 and 3 for β, βα, δγ ∈ (P1)3. Thus intotal there are dimK(P1)i vertices labelled i. The edges of the diagramindicate how these basis elements are transformed under the linear maps ofthe representation. Therefore the edges are labelled by the arrows of thequiver. By convention the maps go from upper to lower rows. Thus, forinstance the edge labelled α represents the map (P1)α : 1 7→ α, whereas all

75

4 Module categories

other basis elements are sent to zero. There are two edges labelled β sinceunder the map (P1)β we have 1 7→ β and α 7→ βα, whereas all other basiselements are sent to zero. If no confusion can arise, the multiplicationindicated on the edges and even the edges itself are omitted. ♦

Proposition 4.22. If A is a basic, finite-dimensional K-algebra and S =e1, . . . , et a complete set of pairwise orthogonal, primitive idempotents,then De1A, . . . ,DetA is a complete list of pairwise non-isomorphic indecom-posable injective A-modules up to isomorphism.

Proof. By Proposition 4.3, the injectives in modA are the projectives in(modA)op ' mod(Aop), that is, the direct summands of Aop. Now thedirect summands of Aop are Aop ·op ei = eiA and therefore the result followsby Remark 4.4 (b).

Remarks 4.23 (a) If Q is the quiver of the algebra A and x ∈ Q0 a vertex,then the injective DexA will be denoted by Ix.

(b) It is useful to translate the description of the injectives into the cate-gorical language. For this, let C be a finite spectroid. Then, to each objectx ∈ C , there corresponds an indecomposable projective object in mod C ,namely Px = C (x,−). The indecomposable injective associated to x isIx = DC (−, x).

Note that here we used tacitly the maps

HomC (α, y) : HomC (x′, y)→ Hom(x, y), β 7→ βα,

HomC (y, α) : HomC (y, x)→ Hom(y, x′), γ 7→ αγ

for x, x′, y ∈ C and α : x→ x′. ♦

Lemma 4.24 (Yoneda’s Lemma). Let A be a finite-dimensional K-algebraand let e1, . . . , et be a complete set of pairwise orthogonal, primitive idem-potents. Then for each A-module M and x = 1, . . . , t the following twomaps

Φx,M : HomA(Px,M)→Mx, f 7→ f(ex)

andΨx,M : HomA(M, Ix)→ DMx, g 7→ g(?)(ex)

are K-linear bijections, where Mx = exM .

76


Proof. Since f(ex) = f(e2x) = exf(ex) we have that Φx,M is well-defined.

Clearly Φx,M is K-linear and it is straightforward to verify that the inverseof Φx,M is given by assigning to m ∈Mx the map Aex →M,aex 7→ aexm.

If m ∈ M then g(m) ∈ Ix = DexA and hence g(m)(ex) ∈ K whichshows that Ψx,M is well defined and K-linear. Again we indicate the in-verse of Ψx,M to show the bijectivity. For ϕ ∈ DMx we define Ψ′x,M (ϕ) ∈HomA(M, Ix) by Ψ′x,M (ϕ)(m) : exA→ K, exr 7→ ϕ(exrm).

Remarks 4.25 (a) If we set M = Py we get HomA(Px, Py) ' (Py)x =exAey and hence to every α ∈ exAey we get a map Pα : Px → Py. Similarlyif we set M = Ix we get that HomA(Ix, Iy) ' D(Ix)y = D(DexA)y =DDexAey ' exAey and to every α ∈ exAey we get a homomorphismIα : Ix → Iy.

In particular, if Q is the quiver of A, then for each arrow α : y → x we gethomomorphisms Pα : Px → Py and Iα : Ix → Iy.

(b) The bijections Φx,M and Ψx,M are functorial in M , that is, if h : M →N is a homomorphism of A-modules, then hxΦx,M = Φx,N HomA(Px, h) andDhxΨx,N = Ψx,M HomA(h, Ix).

Note that it follows from Remark 4.25 (a) that for α ∈ exAey we get twohomomorphisms Pα : Px → Py and Iα : Ix → Iy. Note that Pα =? · α andIα = D(α·?). The bijections Φx,M and Ψx,M are also functorial in x in thesense that for each α ∈ exAey we have MαΦy,M = Φx,M HomA(Pα,M) andDMαΨx,M = Ψy,M HomA(M, Iα). The proof of these functorialities is leftas Exercise 4.5.5. ♦

There are many applications of the Yoneda Lemma and we mention one atonce.

Proposition 4.26. Let A be a finite-dimensional K-algebra with quiver Qand let P (resp. I ) be the full subcategory of mod A composed of allprojective (resp. injective) objects. Then there exists a natural equivalenceνA : P → I defined by νAPx = Ix for each vertex x ∈ Q0 and νAPα = Iαfor each arrow α ∈ Q1.

Proof. In the following sequence the first (and the last) map is given bythe previous lemma and the map in the middle is the natural isomorphism

77

4 Module categories

between a vector space and its double dual.

HomA(Px, Py)→ Py(x) = exAey → DDexAey = DIx(y)→ HomA(Ix, Iy)

Note that all maps are bijections on objects and functorial.

The functor νA is called Nakayama functor.

Remark 4.27. Let A be a basic, finite-dimensionalK-algebra. Then for anycomplete set of pairwise orthogonal, primitive idempotents S = e1, . . . , etthe category C = AS is a spectroid, see Example 3.22. Note that ind P is aspectroid and isomorphic to C op, since HomC (Px, Py) ' C (y, x) = C op(x, y)for all x, y ∈ C . On the other hand, the arrows of the quiver associatedto A via the choice of S correspond to the data rad C / rad C 2, which bythe previous is given entirely by the categorical properties of modA, sincethe indecomposable projectives in modA are completely independent of thechoice of S. This shows finally that the algebra A uniquely determines itsquiver. ♦

Exercises

4.5.1 Proof that an object P of a category C is projective if and only if eachepimorphism ψ : N → P splits, that is, there exists a morphism ϕ : P → N suchthat ψϕ = idP . Note that in case C is additive, it follows then by Exercise 4.3.3that P is isomorphic to a direct summand of N .

4.5.2 Explicitely state the definition of an injective object in modA and draw thecorresponding diagram.

4.5.3 Determine all indecomposable projective A-modules up to isomorphism ifA = KQ, where Q is the two subspace quiver.

4.5.4 Determine all indecomposable projectives and all indecomposable injectivesfor the algebra of Example 3.32.

4.5.5 Prove the four statements of functoriality stated in Remark 4.25 (b).

4.5.6 Show the following description of R = radPx as representation of the quiverof A. Ry = (Px)y = eyAex holds for all y 6= x and Rx is the maximal ideal of thelocal algebra exAex.

4.6 Projective covers and injective hulls

We now study the situation where we have an epimorphism P → M withprojective P . First note that, for all A-modules M there always exists such

78

4.6 Projective covers and injective hulls

an epimorphism: let m1, . . . ,mn be a K-basis of M as K-vector space. Thenthe morphism f : An →M given by f(a1, . . . , an) =

∑ni=1 aimi is surjective

hence epi and An is projective.

A projective cover of an A-module M is an epimorphism f : P → M ,where P is projective such that each endomorphism ρ ∈ EndA(P ) withfρ = f is an isomorphism. The following result shows that a projectivecover is minimal in more than one sense.

Proposition 4.28. Let A be a finite-dimensional K-algebra. Then for anepimorphism f : P →M with P a projective A-module, the following condi-tions are equivalent:

(a) f is a projective cover of M .

(b) P is “closest” to M : for each epimorphism g : Q→M with projectiveQ there exists an epimorphism ψ : Q→ P such that fψ = g.

(c) For each epimorphism g : Q → M with projective Q the module P isisomorphic to a direct summand of Q.

(d) For each epimorphism g : Q→M with projective Q we have dimK P ≤dimK Q.

Proof. Since several implications are based on similar arguments, we willstart by a preparation, valid for each epimorphism g : Q→ M with projec-tive Q. Since P is projective there exists a factorization ϕg : P → Q suchthat gϕg = f and since Q is projective there exists ψg : Q → P such thatfψg = g. Define ρg = ψgϕg ∈ EndA(P ) and observe that fρg = f . Alsorecall from Exercise 4.1.2 that epimorphism in modA is synonymous withsurjective homomorphism.

Assume (a), then ρg is an isomorphism and therefore ψg surjective andsatisfies fψg = g hence (b). Given (b), we get from the projectivity of Pa factorization of idP through the surjective ψg, say idP = ψgh. Then byExercise 4.5.1 the epimorphism ψg splits and P is isomorphic to a directsummand of Q. Clearly (c) implies (d).

Finally assume (d) and let ρ ∈ EndA(P ) be some endomorphism with fρ =f . By Lemma 3.12 there exists an integer n ≥ 1 such that P = Im ρn ⊕Ker ρn. Now f(Ker ρn) = 0 since for each m ∈ Ker ρn we have f(m) =fρn(m) = 0. Thus f restricts to a morphism f ′ : Im ρn → M , which isagain surjective since Im f ′ = Im f = M . Now Im ρn is projective and

79

4 Module categories

therefore dimK Im ρn ≥ dimK P by our assumption. Hence we concludethat Im ρn = P . This implies that ρn and therefore ρ is an isomorphism.This shows (a).

Corollary 4.29. If A is a finite-dimensional K-algebra, then for each A-module M ∈ modA there exists a projective cover. Furthermore, two pro-jective covers are isomorphic, that is, if P →M and Q→M are projectivecovers of M , then P ' Q.

Proof. We have already seen, that for each M ∈ modA there exists anepimorphism An → M for n = dimKM . This shows that there exists anepimorphism f : P → M with projective P and we may assume that thedimension dimK P is minimal among all such epimorphisms. Then by theprevious result f is a projective cover.

If f : P → M and g : Q → M are two projective covers, then by Propo-sition 4.28 there exist an epimorphism ϕ : P → Q with gϕ = f . SincedimK P = dimK Q we conclude that ϕ is an isomorphism. This showsP ' Q.

Remark 4.30. Often, only the projective A-module P is given to indicatea projective cover f : P →M . Due to Corollary 4.29 this makes sense, sinceP is uniquely determined up to isomorphism and any surjective morphismg : P →M will give a projective cover. Moreover, there exists ψ ∈ EndA(P )such that gψ = f and each such ψ will be an isomorphism. ♦

The dual concept of projective cover is called injective hull. Note that wecan dualize all four properties of Proposition 4.28, even the forth one, sincethe dimension does not change under dualization. Therefore we get that foreach A-module M ∈ modA there exists an injective hull M → I and thatI is unique up to isomorphism. Hence modA is an abelian category withenough projectives and injectives, that is, for each object M ∈ modAthere exists a projective cover and an injective hull.

Example 4.31. Let A = K−→A 5, the path algebra of the quiver whose repre-

sentations we studied in Section 2.6. To determine the projective coverof M = [2, 3], we first observe that by Example 4.19, the indecompos-able projective representations are Pi = [i, n] and by Lemma 2.15, we haveHom

K−→A 5

([i, n], [2, 3]) = 0 if i = 1 or i = 4, 5. Note that

ι2,32,5 : [2, 5]→ [2, 3]

80

4.7 Simple modules

is surjective and that ι2,3i,5 is not surjective for i > 2. Hence ι2,32,5 is a projectivecover of M = [2, 3]. ♦

Exercises

4.6.1 Show that if f1 : P1 → M1 and f2 : P2 → M2 are projective covers, thenf1 ⊕ f2 : P1 ⊕ P2 →M1 ⊕M2 is also a projective cover.

4.6.2 Conclude from the previous exercise that if f : P →M1⊕M2 is a projectivecover, then up to isomorphisms P = P1 ⊕ P2 and f = f1 ⊕ f2, where f1 : P1 →M1

and f2 : P2 →M2 are projective covers.

4.6.3 Let Q be the two subspace quiver and M the following indecomposable rep-

resentation: K1−→ K

1←− K. Determine a projective cover of M .

4.7 Simple modules

As in the previous sections, A denotes a finite-dimensional K-algebra. AnA-module M is called simple or irreducible if 0 and M are the onlysubmodules of M .

Example 4.32. Let Pi = Aei be an indecomposable projective A-module.Then the quotient Si = Pi/ radPi is a simple module. As a representation,Si is one-dimensional in the vertex i and zero elsewhere, and (Si)α = 0 foreach arrow α of the quiver of A.

Indeed, it follows from Exercise 4.5.6 that radPi is a complement of Kei inPx viewed as vector spaces. Hence Si = Pi/ radPi is one-dimensional. Thisshows that Si is simple since Si is non-zero and only admits 0 and Si assubmodules. Furthermore dimK(Si)i = 1 and (Si)j = 0 for all j 6= i and(Si)α = 0 since Im(Si)α ⊆ radSi = 0, where the last equations follows fromthe Nakayama Lemma (Proposition 3.20). ♦

Lemma 4.33 (Schur’s Lemma). If S and S′ are two simple A-modules theneach morphism S → S′ is either zero or an isomorphism.

Proof. Let f : S → S′ be a morphism. Then Ker f is a submodule of S. IfKer f = S, then f = 0. Otherwise Ker f = 0 and then f is injective byLemma 4.9. Also, the image f(S) is a submodule of S′ and cannot be zero.Hence f(S) = S′, therefore f is surjective, again by Lemma 4.9, and thusan isomorphism.

81

4 Module categories

The following result shows that the simples of Example 4.32 are all thesimple A-modules.

Proposition 4.34. Let A be a basic, finite-dimensional K-algebra ande1, . . . , et a complete set of pairwise orthogonal, primitive idempotents andPi = Aei the corresponding indecomposable projectives. Then P1/ radP1,. . ., Pt/ radPt is a complete list of pairwise non-isomorphic simple A-modules up to isomorphism.

Proof. Let S be a simple A-module. By Corollary 4.29 there exists a pro-jective cover f : Q → S. Let Q = Q1 ⊕ . . . ⊕ Qm be a decomposition intoindecomposables which are again projective by Lemma 4.16. Let fi : Qi → Sbe the morphism induced by f . Then there exists an index i such that fi 6= 0and Im fi = S follows, since S is simple. This implies that fi : Qi → S isalready an epimorphism. By the minimality of the projective cover, Q = Qiis indecomposable and hence isomomorphic to Ph = Aeh for some h. Thisshows that the projective cover of S if the form f : Ph → S.

Now radPh is a proper submodule of Ph by Nakayama’s Lemma (Proposi-tion 3.20) and therefore Ph/ radPh a non-zero module which is simple, seeExample 4.32. By Remark 3.19, we have f(radPh) ⊆ radS = 0. Thereforef induces a morphism Ph/ radPh → S which is still surjective and thereforeby Schur’s Lemma an isomorphism.

If Q is the quiver of the finite-dimensional K-algebra A then the simplemodule Px/ radPx associated to the vertex x is denoted by Sx.

Note that each non-zero M ∈ modA admits a maximal submodule M1 (M : just take M1 to be a proper submodule of maximal dimension. Thequotient M/M1 must then be a simple module, since otherwise any propersubmodule 0 6= V ( M/M1 could be lifted to M1 ( π−1(V ) ( M , ifπ : M → M/M1 denotes the canonical projection. This can be iteratedresulting in a filtration

0 = M` (M`−1 ( . . . (M1 (M0 = M

of M whose quotients Mi/Mi+1 are simple. Such a filtration is called com-position series or Jordan-Holder filtration and the quotients are calledcomposition factors. We will later see in Exercise 5.1.2 that each simpleSi must occur precisely dimKMi times in such a filtration, where Mi = eiMis the vector space corresponding to the vertex i. This is the analogon ofthe classical Theorem of Jordan-Holder: for any two composition series the

82

4.7 Simple modules

composition factors have the same multiplicities, that is, the compositionfactors are equal up to order and isomorphisms.

Exercises

4.7.1 Show that an A-module M with radM = 0 is semisimple, that is, it isa direct sum of simple A-modules. The best way to see this is to view M as arepresentation of the quiver Q of A. Show that the condition radM = 0 translatesinto Mα = 0 for each arrow α of Q and conclude from this the statement.

4.7.2 Let M be a fixed A-module. Show that if S and S′ are two submodules ofM which are semisimple, then also S + S′ is a semisimple submodule. It thereforemakes sense to speak of “the largest submodule of M which is semisimple”. Thisuniquely defined submodule of M is called socle of M and denoted by socM .

83

4 Module categories

84

Chapter 5

Elements of homologicalalgebra

Homological algebra is one of the most important tools if not the mostimportant of all for the theory of representations of algebras. It gathersinformation of different point of views about its main subject, which is thatof short exact sequences. With this it prepares the basics for the secondstep for understanding the structure of module categories.

5.1 Short exact sequences

First we develop some tools of homological algebra. Although this is nor-mally done in the context of an arbitrary abelian category (with enoughinjectives and projectives) we focus only on module categories reaping thebenefits of working with functions, their arguments and values. Thus, let Abe a finite-dimensional algebra over some algebraically closed field K andmodA the category of finite-dimensional left A-modules.

A short exact sequence in modA is an exact sequence of the form

0→ Xa−→ Y

b−→ Z → 0, (5.1)

with X,Y, Z ∈ modA. Note that this means that a is injective, b is surjectiveand Ker b = Im a. We abbreviate the sequence (5.1) by (a, b) or (a, Y, b) andoften drop the bounding zeros in diagrams. The sequence (5.1) starts inX, ends in Z and Y is its middle term. The two modules X and Zare also called the end terms of the sequence (5.1). For fixed modules

85

5 Elements of homological algebra

X,Z ∈ modA the short exact sequences starting in X and ending in Z donot form a set but a class, which we shall denote by E (Z,X) (that Z ismentioned before X has its own reason, which will become clear later).

We now introduce an equivalence relation among these short exact se-quences: (a, b) ∼ (a′, b′) if there exists an isomorphism ζ such that a′ = ζaand b = b′ζ. The collection of equivalence classes in E (Z,X) will be denotedas Ext1

A(Z,X). As we will see, Ext1A(Z,X) is a set and it carries the struc-

ture of an abelian group, even of a K-vector space. Therefore Ext1A(Z,X)

is called extension group. Note that by definition, the middle terms oftwo equivalent short exact sequences must be isomorphic. That the conversedoes not hold is shown in Example 5.11.

Lemma 5.1. For each X,Z ∈ modA the class Ext1A(Z,X) is a set.

Proof. By Exercise 5.1.1, the middle term Y of each exact sequence (a, b) ∈E (Z,X) has dimension d = dimK X + dimK Z. Therefore, there existsa K-linear bijection ϕ : Y → Kd and we can turn Kd into an A-module,isomorphic to Y , via the multiplication

A×Kd → Kd, (a, v) 7→ a ∗ v = ϕ(aϕ−1(v)).

The short exact sequence (a, Y, b) is therefore equivalent to (ϕa,Kd, bϕ−1).

Thus to determine Ext1A(Z,X) it is enough to consider short exact sequences

of the form (a,Kd, b), where Kd is equipped with some external multiplica-tion A × Kd → Kd. Such short exact sequences form a set since they aregiven as triple (a, b, µ), namely two linear maps a : X → Kd and b : Kd → Zand a bilinear map µ : A×Kd → Kd. Consequently the equivalence classesExt1

A(Z,X) form a set.

The following result shows that it is enough to have a homomorphism be-tween the middle terms in order to establish an equivalence.

Lemma 5.2. Let (a, Y, b), (a′, Y ′, b′) ∈ E (Z,X) be two short exact se-quences. If there exists a homomorphism ζ : Y → Y ′ such that ζa = a′ andb′ζ = b, then ζ is an isomorphism and consequently (a, Y, b) ∼ (a′, Y ′, b′).

Proof. We exhibit a diagram, since it facilitates the orientation.

X

Y

Y ′

Z

*

*H

HHHj

HHHHj

?

a

a′

b

b′

ζ

86


Assume ζ(y) = 0 for some y ∈ Y . Then b(y) = b′ζ(y) = 0 and becauseKer(b) = Im(a) there exists some x ∈ X such that a(x) = y. Now, sincea′(x) = ζa(x) = ζ(y) = 0 it follows from the injectivity of a′ that x = 0 andtherefore y = 0. This shows that ζ is injective. To see the surjectivity of ζlet y′ ∈ Y ′ be some element and set z = b′(y′). Since b is surjective thereexists some y ∈ Y such that b(y) = z. The element y is a good candidatefor a preimage of y′. However ζ(y) 6= y′ is possible, since we only can ensurethat b′ζ(y) = z = b′(y′). Thus we look at the difference y′ − ζ(y) ∈ Y ′ andobserve that b′(y′− ζ(y)) = b′(y′)− b(y) = z− z = 0. Therefore, there existssome x ∈ X such that a′(x) = y′ − ζ(y). Now let t = y + a(x) ∈ Y andcalculate ζ(t) = ζ(y)+ζa(x) = ζ(y)+a′(x) = y′. This shows the surjectivityof ζ.

Remark 5.3. This kind of arguing is called “diagram chasing”, since youfollow the elements as they are pushed along arrows or “sucked back” ifthey lie in some image. Note that in the above proof there is no surpriseand that, once one is on the right track, one has just hang on and follow tothe end.

Each part of algebraic theory has its own taste. In homological algebra it isoften the case that reading a proof is equally affording than finding it afresh.It can be more advantageous for the reader to work the details out on hisor her own, rather than to follow the detailed descriptions written down ina textbook. However, it is also important to see how the mathematics lookswhen it is written down and this is why we have given a detailed accountof the argumentation in the previous case. In the following we often onlyoutline the things to prove and leave it to the reader to do this properly. ♦

In the following, we define a left multiplication on E (Z,X):

HomA(X,X ′)× E (Z,X)→ E (Z,X ′), (h, (a, b)) 7→ h(a, b). (5.2)

Given (a, Y, b) ∈ E (Z,X) and a homomorphism h : X → X ′ we are goingto define a new short exact sequence in E (Z,X ′), denoted by h(a, Y, b), asfollows. First we form the push-out (Y ′, h′, a′) of a and h, as shown in thenext picture.

X Y Z

X ′ Y ′ Z? ?

-

- -

-

a b

a′ ∃!b′

h h′

87


Since ba = 0 and 0h = 0 we have two homomorphisms b : Y → Z and0: X ′ → Z ending in Z and by the universal property of the push-out, seeExercise 4.3.6, there exists a unique b′ : Y ′ → Z such that b′h′ = b andb′a′ = 0. Thus the right square of the preceding picture commutes.

Lemma 5.4. The sequence h(a, b) obtained as explained above is alwaysexact, that is, it belongs to E (Z,X ′).

Proof. Let (a′, b′) = h(a, b). Since b is surjective so must be b′. To see thata′ is injective, we recall from Exercise 4.3.6 that we can assume, withoutloss of generality, that the push-out (Y ′, h′, a′) is obtained as cokernel ψ ofthe map ϕ:

X

ϕ=

[ah

]−−−−→ Y ⊕X ′

ψ=[h′ −a′]−−−−−−−−→ Y ′.

Hence, if a′(x′) = 0 then (0, x′) ∈ Y⊕X ′ satisfies ψ(0, x′) = h′(0)−a′(x′) = 0.Since modA is abelian, we have Ker(ψ) = Im(ϕ) = ϕ(X) and therefore thereexists an element x ∈ X such that 0 = a(x) and x′ = h(x). The injectivityof a implies x = 0 and x′ = 0 follows. This shows that a′ is injective.

Since b′a′ = 0 we have Im a′ ⊆ Ker b′. Now, suppose that some elementy′ = ψ(y, x′) = h′(y)−a′(x′) lies in Ker b′. Then 0 = b′(y′) = b(y) shows thaty = a(x) for some x ∈ X since (a, b) is exact. Hence y′ = h′a(x)− a′(x′) =a′(h(x) − x′) ∈ Im(a′). Thus, we have shown that Ker b′ = Im a′. Hence(a′, b′) is a again a short exact sequence.

Thus, we have defined the left multiplication (5.2).

Proposition 5.5. The left multiplication defined in (5.2) induces a multi-plication

HomA(X,X ′)× Ext1A(Z,X)→ Ext1

A(Z,X ′), (h, ε) 7→ hε (5.3)

which is associative in the sense that for each h ∈ HomA(X,X ′) and eachk ∈ Hom(X ′, X ′′) we have k(hε) = (kh)ε.

Proof. We have to show that, if (c, T, d) is equivalent to (a, Y, b), thenh(a, Y, b) is equivalent to h(c, T, d). Let (a′, Y ′, b′) = h(a, Y, b) as aboveand similarly (c′, T ′, d′) = h(c, T, d) with h′′ : T → T ′ the correspondinghomomorphism of the middle terms.

Since (a, b) ∼ (c, d) there exists an isomorphism ξ : Y → T such that c = ξaand dξ = b. Hence (h′′ξ)a = h′′c = c′h. So, by the push-out property of

88


Y ′, there exists a unique homomorphism ρ : Y ′ → T ′ with h′′ξ = ρh′ andc′ = ρa′. Similarly, (h′ξ−1)c = h′a = a′h and by the push-out property ofT ′, there exists a homomorphism σ : T ′ → Y ′ such that (h′ξ−1) = σh′′ anda′ = σc′. Hence (σρ)a′ = a′ and (σρ)h′ = h′. But only the identity map idY ′

has this property. Thus σρ = idY ′ and similarly one sees that ρσ = idT ′ .Therefore we have (a′, b′) ∼ (c′, d′).

This shows, that (5.2) induces a map (5.3). It remains to verify the statedassociativity.

Let (Y ′, h′, a′) be a push-out of a and h, next let (Y ′′, k′, a′′) be a push-outof a′ and k and finally let (V, `′, e) be a push-out of a and ` = kh, see thenext picture for illustration. Then (k′h′)a = k′a′h = a′′(kh), thus by theuniversal property of (V, `′, e) there exists a unique ϕ : V → Y ′′ such thatϕ`′ = k′h′ and ϕe = a′′. By the same property f : V → Z is uniquelydetermined by fe = 0 and f`′ = b. Therefore, since b′′ϕe = b′′a′′ = 0 andb′′ϕ`′ = b′′k′h′ = b′h′ = b we must have f = b′′ϕ.

X

X ′

X ′′

Y

Y ′

Y ′′ Z

Z

Z

X ′′ V Z

- -

- -

- -

- -

? ?

? ?

a b

a′ b′

a′′ b′′

e f

h

k

h′

k′` = kh `′

ϕ

By Lemma 5.2 the two sequences (e, V, f) and (a′′, Y ′′, b′′) are equivalent.

The construction can be dualized: given a short exact sequence (a, b) ∈E (X,Z) and g ∈ HomA(Z ′, Z) we can form the pull-back (Y ′, g′, b′) of band g and obtain an exact sequence (a′, b′), which is denoted by (a, b)g.Thus a right multiplication is defined. The proof of the following resultis obtained by dualizing the proof of Proposition 5.5.

Proposition 5.6. The right multipilication of short exact sequences inducesa well defined right multiplication

Ext1A(Z,X)×HomA(Z ′, Z)→ Ext1

A(Z ′, X), (ε, g) 7→ εg. (5.4)

89


Furthermore, this right multiplication is associative in the sense that(εg)g′ = ε(gg′) for each g′ ∈ HomA(Z ′′, Z ′).

The proof of the following compatibility result is left as Exercise 5.1.3 to thereader.

Lemma 5.7. The left multiplication (5.3) is compatible with the right mul-tiplication (5.4), that is, for all ε ∈ Ext1

A(Z,X), all h ∈ HomA(X,X ′) andall g ∈ Hom(Z ′, Z) we have (hε)g = h(εg).

Exercises

5.1.1 Show that for a short exact sequence (5.1) the following dimension formulaholds: dimK Y = dimK X + dimK Z. Improve this by showing that dimY =dimX + dimZ.

5.1.2 Use the previous exercise and the description of the simple modules of Ex-ample 4.32 to show that in a Jordan-Holder filtration of an A-module M the simpleAei/ radAei occurs precisely dimKMi times.

5.1.3 Prove Lemma 5.7. Draw a diagram of the picture and label the homomor-phisms appearing in it. Then work your way through it always using the propertiesof push-outs and pull-backs.

5.1.4 Show that two consecutive homomorphisms Xa−→ Y

b−→ Z form a short exactsequence if and only if a is a kernel of b and b is a cokernel of a.

5.1.5 Show that if 0 → Ua−→ V

b−→ W → 0 is a short exact sequence of vectorspaces, then dualization over the ground field yields a short exact sequence 0 →DW

Db−−→ DVDa−−→ DU → 0.

5.1.6 Prove that the short exact sequence 0 → radM → M → M/ radM → 0 isdual to 0→ socM →M →M/ socM → 0.

5.2 The Baer sum

In the previous section we have gathered all the necessary preliminaries todefine a binary operation in the set Ext1

A(Z,X), which will turn Ext1A(Z,X)

into an abelian group. This operation is called Baer sum. For two shortexact sequences (a, b) and (c, d) in E (Z,X), we define their direct sumby (a, b) ⊕ (c, d) = (a ⊕ c, b ⊕ d) ∈ E (Z ⊕ Z,X ⊕ X). Then, using the

90

5.2 The Baer sum

homomorphisms ΣX : X⊕X → X, (x, x′) 7→ x+x′ and ∆Z : Z → Z⊕Z, z 7→(z, z), define

(a, b) + (c, d) = ΣX(a⊕ c, b⊕ d)∆Z , (5.5)

which is again a short exact sequence in E (Z,X).

In order to see that the binary operation (5.5) satisfies good properties, weneed a rather technical result:

Lemma 5.8. For each (a, b) ∈ E (Z,X) we have ∆X(a, b) ∼ (a⊕a, b⊕b)∆Z

and ΣX(a⊕ a, b⊕ b) ∼ (a, b)ΣZ .

Proof. We first prove that ∆X(a, b) is equivalent to the lower row of thenext picture.

X Y Z

X ⊕X X ⊕ Y Z? ?

- -

- -

a b

a′ =

[−1 1a 0

]b′ =

[0 b

]∆X =

[11

]∆′ =

[01

]

We start by verifying that (X⊕Y,∆′, a′) is a push-out of a and ∆X . Let Mbe an A-module and further let g =

[g1 g2

]: X ⊕X →M and h : Y →M

be two homomorphisms such that g∆X = ha, that is, g1+g2 = ha. Then thehomomorphism f =

[g2 h

]: X ⊕ Y →M satisfies fa′ =

[−g2 + ha g2

]=[

g1 g2

]= g and clearly f∆′ = h. To see the uniqueness of f we assume

that the homomorphism f ′ =[f ′1 f ′2

]: X ⊕ Y → M satisfies f ′a′ = g and

f ′∆′ = h. Then the first equation implies f ′1 = g2 and the second f ′2 = hshowing f ′ = f . Hence (X ⊕ Y,∆′, a′) is indeed a push-out of a and ∆X .Since b = b′∆′ and b′a′ = 0 we conclude that ∆X(a, b) ∼ (a′, b′).

Now we prove that (a ⊕ a, b ⊕ b)∆Z ∼ (a′, b′). For this, we consider thefollowing picture.

X ⊕X Y ⊕ Y Z ⊕ Z

X ⊕X X ⊕ Y Z

6 6

- -

- -

a⊕ a =

[a 00 a

]b⊕ b =

[b 00 b

]

a′ =

[−1 1a 0

]b′ =

[0 b

]∆Z =

[11

]∆′′ =

[0 1a 1

]

91


The proof that (X ⊕Y,∆′′, b′) is a pull-back of b⊕ b and ∆Z is quite similarto the one above and we therefore leave it to the reader. A direct calculationshows that ∆′′a′ = a⊕ a and b′a′ = 0, thus (a⊕ a, b⊕ b)∆Z ∼ (a′, b′). Thesecond equivalence follows by dualization.

Theorem 5.9. The set Ext1A(Z,X) together with the Baer sum, defined

in (5.5), is an abelian group. Furthermore, the left- and right-multiplicationdefined in Propositions 5.5 and 5.6 distribute over the Baer sum. That is, forall h, g ∈ Hom(X,X ′) and all ε ∈ Ext1

A(Z,X) the equality hε+gε = (h+g)εholds and an analog equation is valid for the right multiplication. As aconsequence Ext1

A(Z,X) is a K-vector space.

Proof. The proof is done in several steps.

(i) The Baer sum is commutative. Clearly (a⊕ c, b⊕ d) and (c⊕ a, d⊕ b)are equivalent short exact sequences, hence by Proposition 5.5 and 5.6 weget ΣX(a⊕ c, b⊕ d)∆Z ∼ ΣX(c⊕ a, d⊕ b)∆Z .

(ii) The Baer sum is associative. Observe that

[(a, b) + (c, d)] + (e, f) = ΣX(ΣX ⊕ idX)S (∆Z ⊕ idZ)∆Z ,

(a, b) + [(c, d) + (e, f)] = ΣX(idX ⊕ΣX)S (idZ ⊕∆Z)∆Z ,

where S = (a⊕ c⊕ e, b⊕ d⊕ f). The associativity follows thus from

ΣX(ΣX ⊕ idX) = ΣX(idX ⊕ ΣX) : X ⊕X ⊕X → X

(∆Z ⊕ idZ)∆Z = (idZ ⊕∆Z)∆Z : Z → Z ⊕ Z ⊕ Z.

(iii) The split short exact sequence

ε0 : X

[10

]−−→ X ⊕ Z

[0 1]−−−−→ Z

is a neutral element of the Baer sum. Let (a, Y, b) ∈ E (Z,X). We first showthat the lower sequence of the following diagram is indeed ΣX(ε0 ⊕ (a, b)).

92

5.2 The Baer sum

X ⊕X X ⊕ Z ⊕ Y Z ⊕ Z

X Z ⊕ Y Z ⊕ Z? ?

- -

- -

a′ =

1 00 00 a

b′ =

[0 1 00 0 b

]

a′′ =

[0a

]b′′ =

[1 00 b

]ΣX =

[1 1

]Σ′ =

[0 1 0a 0 1

]

A direct inspection shows that the diagram is commutative. We nowshow that (Z ⊕ Y,Σ′, a′′) satisfies the universal property of a push-out ofa′ and ΣX . For this let M be an A-module and let g : X → M andh =

[h1 h2 h3

]: X ⊕ Z ⊕ Y → M be two homomorphisms such that

ha′ = ΣXg. This last equation is:[h1 h3a

]=[g g

]and hence h1 = g = h3a. Let f =

[h2 h3

]: Z ⊕ Y → M and verify

directly that fΣ′ =[h3a h2 h3

]= h and fa′′ = h3a = h1 = g. This

shows that f satisfies the required factorization property. Moreover for anyf ′ =

[f ′1 f ′2

]: Z⊕Y →M with f ′Σ′ = h and f ′a′′ = g, the former equation

implies f ′1 = h2 and f ′2 = h3, that is, f ′ = f . This shows that (Z⊕Y,Σ′, a′′)is a push-out of a′ and ΣX . Furthermore, we have b′′Σ′ = b′ and b′′a′′ = 0 andit follows therefore from the uniqueness of the push-out, see Exercise 4.3.5for the dual statement, that (a′′, b′′) ∼ ΣX(ε0 ⊕ (a, b)).

We now show that the lower sequence of the next picture is ε0 + (a, b) =ΣX(ε0 ⊕ (a, b))∆Z .

X Z ⊕ Y Z ⊕ Z

X Y Z

6 6

- -

- -

a b

a′′ =

[0a

]b′′ =

[1 00 b

]

∆Z =

[11

]∆′ =

[b1

]

First of all, the diagram commutes. We have to verify the universal pull-back property of (Y,∆′, b). Thus assume that M is an A-module and further

93


that g =

[g1

g2

]: M → Z ⊕ Y and h : M → Z are homomorphisms such that

b′′g = ∆Zh. The last equation means g1 = h = bg2 which can be interpretedas follows: g2 : M → Y is the factorization which satisfies ∆′g2 = g andbg2 = h. Moreover, this is the unique factorization with this property sincefor any f : M → Y the equation ∆′f = g implies f = g2. Thus, the lowerrow of the previous picture is indeed ε0 + (a, b). This shows that ε0 is aneutral element in Ext1

A(Z,X).

(iv) The left and right multiplication are distributive over the Baer sum.Let h, g ∈ Hom(X,X ′) and ε ∈ Ext1

A(Z,X). We have to show that hε+gε =(h + g)ε. For this, observe that h + g = ΣX(h ⊕ g)∆X (usual compositionof homomorphisms) and (h⊕ g)ε = hε⊕ gε and therefore, using Lemma 5.8in equation (∗), we get

(h+ g)ε = ΣX(h⊕ g)∆Xε(∗)= ΣX(h⊕ g)ε∆Z = ΣX(hε⊕ gε)∆Z = hε+ gε.

The distributivity with respect to right multiplication is obtained dually.

(v) Existence of additive inverses. The following equalities, where λ ∈ K,are subject of Exercise 5.2.1:

λ(a, b) =

( 1λa, b), if λ 6= 0,

ε0, if λ = 0.(5.6)

Therefore the additive inverse of (a, b) is (−a, b) since

(a, b) + (−a, b) = 1(a, b) + (−1)(a, b) = (1 + (−1))(a, b) = 0(a, b) = ε0,

where we used the distributivity (iv). This concludes the proof.

Remark 5.10. We can rephrase the statement of Theorem 5.9 as follows:The functor Ext1

A(Z,−) : modA → vec is a covariant K-linear functor foreach A-module Z and Ext1

A(−, X) : modA→ vec is a contravariantK-linearfunctor for each A-module X. Since the left and right action are compatibleby Lemma 5.7 we even get that Ext1

A is a bifunctor, that is, a functor

modAop ×modA→ vec, (Z,X) 7→ Ext1A(Z,X),

where C ×D , for two categories C and D , denotes the product category,whose objects are pairs (C,D) with C ∈ C and D ∈ D and whose morphisms(C,D)→ (C ′, D′) are pairs (f, g) with f ∈ C (C,C ′) and g ∈ D(D,D′) withthe obvious composition and identity morphisms. ♦

94

5.2 The Baer sum

It is not easy to determine the extension groups just using the definition,except in simple examples of low dimensions.

Example 5.11. Let A = KQ be the Kronecker algebra, that is the pathalgebra of the quiver Q with two vertices 1 and 2 and two arrows α, β : 1→ 2.

Let us calculate Ext1A(S1, S2), where Si denotes the simple A-modules asso-

ciated to the vertex i, see Example 4.32. Using Exercises 6.3.1 and 5.1.1, weknow that the middle term Y of each short exact sequence 0→ S2 → Y →S1 → 0 has, as representation of Q, dimension one in each of the vertices.Up to an isomorphism of A-modules, or equivalently an equivalence of shortexact sequences, we can assume that Y1 = Y2 = K. Then Y is given bytwo linear maps Yα = [γ] and Yβ = [δ] for some γ, δ ∈ K. We abbreviatethe representation Y with these two linear maps by (γ, δ) ∈ K2. Thus, wehave to consider short exact sequences of representations as shown in thefollowing picture.

K

0

K

K

0

K

?? ?? ??- -

- -

a

b

γ δ

Each such exact sequence is given by four K-linear maps K → K, that is,by four scalars a, b, γ, δ ∈ K, where a 6= 0 and b 6= 0. Note that (γ, δ) and(λγ, λδ) for each λ ∈ K∗ = K \ 0 define isomorphic representations. Thussetting ϕ1 = b and ϕ2 = 1

a we get an isomorphism ϕ : (γ, δ) → ( γab ,δab) =

(γ′, δ′) and have then an equivalent short exact sequence, which denote asε(γ′, δ′):

K

0

K

K

0

K

?? ?? ??- -

- -

1

1

γ′ δ′ (5.7)

Putting all this together, we get that each equivalence class E (S1, S2) con-tains an element of the form (5.7). Note that this element is unique. Hencewe see that Ext1

A(S1, S2) has dimension two and can be considered as theset ε(α, β) | α, β ∈ K.The sequence ε(0, 0) is special: it is the split exact sequence. For all otherpairs (γ, δ) ∈ K2 the set ε(λγ, λδ) | λ 6= 0 consists of pairwise non-equivalent short exact sequences which have isomorphic middle terms. InExercise 5.2.2 the additive behavior is studied further. ♦

95


Exercises

5.2.1 Show the equalities stated in the equations (5.6).

5.2.2 Show that in Example 5.11 the following additivity holds: ε(α, β)+ε(γ, δ) =ε(α+ γ, β + δ).

5.2.3 Let ε = (Xa−→ Y

b−→ Z) be a short exact sequence and g : Z ′ → Z somehomomorphism. Use Exercise 4.3.7 to show that εg = 0 if and only if g factorsthrough b, that is, there exists some ξ : Z ′ → Y such that g = bξ.

5.3 Resolutions

Example 5.11 shows that for a finite-dimensional K-algebra A and twoA-modules X,Z it is a difficult problem to compute the extension groupExt1

A(Z,X) following the presented definition: we had to consider all possi-ble middle terms Y , from which we only know the dimension (or dimensionvector, if the modules are considered as representations of a quiver, see Ex-ercise 6.3.1) and then all possible extensions X → Y → Z. Afterwards, wehad to describe the equivalence classes. That seems like a lot of work.

Therefore we want to present a different approach which is far better forpractical purpose and even allows a generalization.

A projective resolution of an A-module M is an exact sequence

· · · ϕ4−→ Q3ϕ3−→ Q2

ϕ2−→ Q1ϕ1−→ Q0 (5.8)

where each object Qi is projective and Cokerϕ1 = M . Dually, an injectiveresolution of M is an exact sequence

J0ψ1−→ J1

ψ2−→ J2ψ3−→ · · ·

where each Jj is injective and Kerψ1 = M .

Remark 5.12. It is common to complete (5.8) to an exact sequence

· · · ϕ4−→ Q3ϕ3−→ Q2

ϕ2−→ Q1ϕ1−→ Q0

ϕ0−→M → 0

and even to call this the projective resolution of M . Similarly an injectiveresolution can be completed with ψ0 : M → J0. ♦

Projective resolutions are often easy to calculate.

96

5.3 Resolutions

Example 5.13. Let A = KQ/I where Q =−→A 5 and the ideal I is generated

by α3α2α1 and α4α3α2, see the next picture, where the two zero relationsare indicated by dotted arcs.

1 2 3 4 5α1 α2 α3 α4

- - - -

We calculate the projective resolution of S1. For this we start with theprojective cover ϕ0 : P1 = [1, 3] → S1. The kernel of ϕ0 is [2, 3] and theprojective cover of [2, 3] is P2 = [2, 4]. This yields ϕ1 = Pα1 : P2 → P1.Proceeding this way we obtain the projective resolution as shown in the toprow of the next picture.

[1, 1][1, 3]

[2, 3]

[2, 4]

[4, 4]

[4, 5]

[5, 5]

[5, 5] ----

@@@R

@@@R

@@@R

ϕ0ϕ1ϕ2ϕ3

||||||Kerϕ0Kerϕ1Kerϕ2

Note that the projective resolution of S1 can be rewritten as 0 → P5Pα4−−→

P4Pα3α2−−−−→ P2

Pα1−−→ P1. ♦

We now want to use these resolutions to the compute extensions groups.Therefore, let Z be an A-module and let

· · · ϕ4−→ Q3ϕ3−→ Q2

ϕ2−→ Q1ϕ1−→ Q0 (5.9)

be a projective resolution of Z with Cokerϕ1 = ϕ0 : Q0 →M .

For a homomorphism h : Q1 → X, we consider the push-out (Yh, ah, h′) of h

and ϕ1, see next picture.

Q2

X

Q1

Yh

Q0

Z

Z

- -

- --

? ?

ah bh

ϕ2 ϕ1 ϕ0

h h′(5.10)

Since ϕ0ϕ1 = 0 = 0h it follows from the universal property of the push-out(Yh, ah, h

′) that there exists a unique homomorphism bh : Yh → Z such thatbhah = 0 and bhh

′ = ϕ0. Thus the diagram commutes.

97


Lemma 5.14. The sequence (ah, bh) as constructed above is a short exactsequence if and only if ah is injective and this happens if and only if hϕ2 = 0.

Proof. The exactness of Xah−→ Yh

bh−→ Z → 0 follows precisely as in theproof of Lemma 5.4. Therefore (ah, bh) is a short exact sequence if an onlyif ah is injective.

To see the second equivalence, observe that if x = hϕ2(p) 6= 0 for somep ∈ Q2 with p 6= 0, then 0 = h′ϕ1ϕ2(p) = ah(x) showing that ah is notinjective. Conversely, if ah(x) = 0 for some x 6= 0, then there exists anelement q ∈ Q1 with h(q) = x and ϕ1(q) = 0 showing that q = ϕ2(p) forsome p ∈ Q2 and hϕ2 6= 0.

Recall from Remark 4.23 (b) that we denoted by HomA(ϕ2, X) the map

HomA(ϕ2, X) : HomA(Q1, X)→ HomA(Q2, X), h 7→ hϕ2.

Hence, the above shows that for any homomorphism h in Ker HomA(ϕ2, X)some short exact sequence (ah, bh) is obtained. We denote by ε(h) theequivalence class of (ah, bh) in Ext1

A(Z,X). Hence, the construction yieldsa map

ε : Ker HomA(ϕ2, X)→ Ext1A(Z,X), h 7→ ε(h). (5.11)

Note that we do not distinguish in our notation between the short exactsequence and its equivalence class.

Proposition 5.15. Let X,Z be A-modules and let (5.9) be a projectiveresolution of Z. Then the map (5.11) induces a K-linear bijection

Ker HomA(ϕ2, X)/ Im HomA(ϕ1, X)→ Ext1A(Z,X). (5.12)

which is functorial in X.

Proof. The proof is rather long and therefore separated in several steps.

(i) The map (5.11) is surjective. Let (a, Y, b) ∈ E (Z,X) be given. Theprojectivity of Q0 implies that there exists h′′ : Q0 → Y such that bh′′ = ϕ0.Since 0 = ϕ0ϕ1 = bh′′ϕ1, the homomorphism h′′ϕ1 factors over Ker b = Im aand therefore over a, that is, there exists h : Q1 → X with ah = h′′ϕ1.Let ε(h) = (ah, Yh, bh) be the short exact sequence obtained as above andh′ : Q0 → Yh the homomorphism which satisfies h′ϕ1 = ahh and bhh

′ = ϕ0.

We show that (a, Y, b) is equivalent to (ah, Yh, bh). Since h′′ϕ1 = ah, thereexists a unique homomorphism ϑ : Yh → Y such that h′′ = ϑh′ and a = ϑah.

98

5.3 Resolutions

Now, bh and bϑ are both homomorphisms with bhah = 0 = ba = (bϑ)ah andbhh′ = ϕ0 = bh′′ = (bϑ)h′, thus it follows from the uniqueness stated in the

universal property of the push-out (Yh, ah, h′) that bh = bϑ. This shows that

(a, b) ∼ (ah, bh).

(ii) The map (5.11) satisfies ε(h) = ε(h+ ξϕ1) for each h ∈ HomA(Q1, X)and each ξ ∈ HomA(Q0, X). Let g = h + ξϕ1. We have to show that(ah, Yh, bh) ∼ (ag, Yg, bg). Note that we have homomorphisms ag : X → Ygand g′ : Q0 → Yg satisfying agg = g′ϕ1 and therefore agh = ag(g − ξϕ1) =(g′ − agξ)ϕ1. By the universal property of the push-out (Yg, ag, g

′) thereexists a unique homomorphism σ : Yg → Yh such that σag = ah and σg′ =h′. We show that bhσ = bg holds: indeed, by the universal property thehomomorphism bg is unique with the properties bgag = 0 and bgg

′ = ϕ0.Since (bhσ)ag = bhah = 0 and (bhσ)g′ = bhh

′ = ϕ it follows that bhσ = bgand therefore (ah, bh) ∼ (ag, bg).

(iii) The map (5.12) is a bijection. By (i) we already know that themap is surjective. It remains to show that ε(h) = ε(g) implies h − g ∈Im HomA(ϕ1, X). Again we denote (ah, Yh, bh) = ε(h) and (ag, Yg, bg) = ε(g)and suppose that there is an isomorphism σ : Yh → Yg such that ag = σahand bh = bgσ. Then ah(h − g) = ahh − σ−1agg = (h′ − σ−1g′)ϕ1. Sincebh(h′ − σ−1g′) = ϕ0 − bgg′ = 0 there exists a homomorphism ξ : Q0 → Xsuch that ahξ = (h′−σ−1g′) and hence ah(h−g) = ahξϕ1. By the injectivityof ah, we conclude h− g = ξϕ1 ∈ Im HomA(ϕ1, X).

(iv) The bijection (5.12) is K-linear. This follows from (5.6) in the proofof Theorem 5.9 together with the similar equality

(a, b)λ =

(a, b 1

λ), if λ 6= 0,

ε0, if λ = 0

and the elementary observation that (a, b 1λ) ∼ ( 1

λa, b).

(v) The bijection (5.12) is functorial in X. For given h ∈ Ker HomA(ϕ2, X)and ξ : X → X ′, we have ξh ∈ Ker HomA(ϕ2, X

′) and it follows ξε(h) =ε(ξh) by the push-out property by the same arguments as the associativitywas shown in Proposition 5.5.

Remark 5.16. The result shows that Ker HomA(ϕ2, X)/ Im HomA(ϕ1, X)is independent of the choice of a projective resolution of Z. Even more, onecan show that this quotient is functorial in Z. Since we will not use this, weleave the proof to the interested reader. ♦

99


The dual construction starts with an injective resolution of the A-moduleX:

J0ψ1−→ J1

ψ2−→ J2ψ3−→ · · ·

and using pull backs along homomorphisms Z → J1 the following bijectionis obtained

Ker HomA(Z,ψ2)/ Im HomA(Z,ψ1)→ Ext1A(Z,X). (5.13)

In the forthcoming we will not distinguish between any of these three de-scriptions and uniformly denote them by Ext1

A(Z,X).

Example 5.17. Continuing Example 5.13, we can calculate the extensiongroup Ext1

A(S1,M) for various indecomposable A-modules M = [i, j]. ByProposition 5.15, we only have Ext1

A(S1,M) 6= 0 if HomA(P2,M) 6= 0, henceby Yoneda’s Lemma 4.24 if M2 6= 0. Furthermore, we only need to considerh : P2 → M if hϕ2 = 0. We conclude i ≤ 2 ≤ j. Since ϕ2 = Pα3α2 and allindecomposable representations are of the form [i, j] we must have M4 = 0,thus j ≤ 4. It remains to determine which homomorphisms h factor throughϕ1 = Pα1 . Clearly this is the case if and only if i = 1. Thus we concludethat Ext1

A(S1, [i, j]) 6= 0 if and only if i = 2 and j = 2, 3. ♦

The approach with resolutions has the advantage that we can define higherextension groups by setting

ExtiA(Z,X) = Ker HomA(ϕi+1, X)/ Im HomA(ϕi, X).

The length of the projective resolution (5.8) of M is just the maximalnumber i such that Qi 6= 0 if such a number exists and otherwise infinite.We define the projective dimension pdimM of an A-module M as theminimal length of any possible projective resolution of M . Hence pdimMis either a non-negative integer or ∞.

Similarly the injective dimension of M is the minimal length occurring inthe injective resolutions of M , it is denoted by idimM . The supremum of theprojective dimensions of all A-modules is called the global dimension. Wedenote the global dimension of A by gldimA and observe that it coincideswith the supremum of all injective dimensions of all A-modules, since it isjust the supremum of all i such that there exists some A-modules M and Nwith Exti(M,N) 6= 0.

The finitistic dimension findimA of the algebra A is by definition thesupremum of all projective dimensions which are finite, that is,

findimA = suppdimM |M ∈ modA, pdimM <∞.

100

5.4 Long exact sequences

There is an interesting open problem about the finitistic dimension.

Conjecture 5.18 (Finitistic dimension conjecture). The finitistic dimen-sion of each finite-dimensional K-algebra is finite.

Exercises

5.3.1 Let A be the algebra considered in Example 5.13. Determine the projectiveresolutions of all simple A-modules Si for i > 1.

5.3.2 Prove that the dimension of the space Ext1A(Si, Sj) is precisely the number

of arrows from j to i in the quiver of A.

5.3.3 Show that by definition Ext0A(M,X) is isomorphic to HomA(M,X) for any

two A-modules M,X.

5.3.4 Show that an A-module M is projective if and only if Ext1A(M,N) = 0 holds

for all A-modules N .

5.3.5 Show that the projective dimension of an A-module M is precisely the max-imal i such that ExtiA(M,A) 6= 0.


As usual let A be a finite-dimensional K-algebra. Fix a short exact sequence

Xa−→ Y

b−→ Z. Then for any A-module M we can compare the three exten-sion groups ExtiA(M,X), ExtiA(M,Y ) and ExtiA(M,Z). For this, we fix aprojective resolution of M

· · ·Q3ϕ3−→ Q2

ϕ2−→ Q1ϕ1−→ Q0.

Note that for each i ≥ 0 we have a sequence

Hom(Qi, X)Hom(Qi,a)−−−−−−−→ Hom(Qi, Y )

Hom(Qi,b)−−−−−−→ Hom(Qi, Z). (5.14)

Note that we have dropped the subscripts indicating the algebra A to sim-plify the notation. If f ∈ Ker Hom(ϕi+1, X), that is, f : Qi → X satisfiesfϕi+1 = 0, then afϕi+1 = 0, that is af ∈ Ker Hom(ϕi+1, Y ). Hence, thesequence (5.14) induces a sequence

Ker Hom(ϕi+1, X)→ Ker Hom(ϕi+1, Y )→ Ker Hom(ϕi+1, Z).

101


Similarly, if f ∈ Im Hom(ϕi, X), that is, f = hϕi for some h ∈Hom(Qi−1, X), then af = ahϕi ∈ Im Hom(ϕi, Y ). This shows that thesequence (5.14) induces also the sequence

Im Hom(ϕi, X)→ Im Hom(ϕi, Y )→ Im Hom(ϕi, Z).

Thus, we get a sequence of the quotients which by Proposition 5.15 is

Exti(M,X)Exti(M,a)−−−−−−→ Exti(M,Y )

Exti(M,b)−−−−−−→ Exti(M,Z). (5.15)

We now present a construction which relates Exti(M,Z) with Exti+1(M,X).For this we assume that ε ∈ Exti(M,Z) is represented by f : Qi → Z withfϕi+1 = 0. Since b : Y → Z is surjective and Qi projective there exists afactorization ` : Qi → Y such that f = b`.

X

Qi+1

Y

Qi

Z- -

-

? ?

@@@@@R

a b

ϕi+1

fh `

Note that fϕi+1 = 0 implies that b(`ϕi+1) = 0 and since a is a kernel ofb, see Exercise 5.1.4, there exists a factorization h : Qi+1 → X such that`ϕi+1 = ah. Hence ahϕi+2 = `ϕi+1ϕi+2 = 0 follows and since a is injectivethis implies hϕi+2 = 0, that is, h ∈ Ker Hom(ϕi+2, X) ⊆ Hom(Qi+1, X). Bytaking the equivalence class of h we get an element of Exti+1(M,X) whichis denoted by δi(ε).

Lemma 5.19. The construction explained above yields a well-defined mapδi : ExtiA(M,Z)→ Exti+1

A (M,X) for each i ≥ 0.

Proof. We verify that the equivalence class of h does not depend on thepossible choices. If f ′ represents the same equivalence class as f , then f ′ =f + gϕi for some g : Qi−1 → Z. Now let `′ : Qi → Y be chosen such thatb`′ = f ′ and let h′ : Qi+1 → X be the unique factorization such that ah′ =`′ϕi+1. We have to show that h and h′ define the same equivalence class inExti+1(M,X).

Again, since Qi−1 is projective and b surjective, we get m : Qi−1 → Y suchthat bm = g. Hence bmϕi = gϕi = f ′ − f = b(`′ − `). Therefore b(`′ −` − mϕi) = 0 and this implies that there exists k : Qi → X such that

102


`′−`−mϕi = ak. Consequently a(h′−h) = (`′−`)ϕi+1 = (ak+mϕi)ϕi+1 =akϕi+1. By the injectivity of a it follows that h′−h = kϕi+1, that is, h′ andh define the same class in Exti+1(M,X).

The maps δi are called connecting maps. The next result is very importantfor dealing with extension groups.

Theorem 5.20. Let Xa−→ Y

b−→ Z be a short exact sequence of A-modules.Then for each A-module M the infinite sequence

HomA(M,X) HomA(M,Y ) HomA(M,Z)

Ext1A(M,X) Ext1

A(M,Y ) Ext1A(M,Z)

Ext2A(M,X)

Exti−1A (M,Z)

ExtiA(M,X) ExtiA(M,Y ) ExtiA(M,Z)

Exti+1A (M,X)

0

-

- --

- -

-

- - -

-

-

-

· · ·

· · ·

· · ·

δi

δi−1

δ1

δ0-

is exact, where all the horizontal maps are as in (5.15).

Proof. Note, that we already have used Exercise 5.3.3 to replace Ext0 byHom. In the sequel we will drop again the subscript indicating the algebraA. The proof is done in three steps.

(i) The sequence (5.15) is exact for each i ≥ 0. Let ε ∈ Exti(M,Y ) berepresented by f ∈ Hom(Qi, Y ) with fϕi+1 = 0. Now Exti(M, b)(ε) isrepresented by the homomorphism bf , thus Exti(M, b)(ε) = 0 means thatbf represents the equivalence class which contains the zero map. In otherwords we have bf = mϕi for some m : Qi−1 → Z. Since Qi−1 is projectiveand b surjective, there exists ` : Qi−1 → Y such that b` = m. This impliesb(f − `ϕi) = 0 and consequently there exists some h : Qi → X such thatf − `ϕi = ah, that is, f and ah represent the same equivalence class inExti(M,Y ). If η is the equivalence class of h in Exti(M,X), then ε = aη.This shows Ker Exti(M, b) ⊆ Im Exti(M,a). The other inclusion followsdirectly from b(aη) = (ba)η = 0η = 0.

103


(ii) For each i ≥ 0, the sequence

Exti(M,Y )Exti(M,b)−−−−−−→ Exti(M,Z)

δi−→ Exti+1(M,X)

is exact. Let ε ∈ Exti(M,Z) be represented by f ∈ Hom(Qi, Z) withfϕi+1 = 0. Now, if ε ∈ Im Exti(M, b) then f = b`′ + mϕi for some `′ ∈Hom(Qi, Y ) with `′ϕi+1 = 0 and some m ∈ Hom(Qi−1, Z). By Lemma 5.19we have δi(f) = δi(b`) and by construction δi(b`) = 0 follows since thefactorization h : Qi+1 → X with ah = `ϕi+1 = 0 can be chosen to be h = 0.This shows Im Exti(M, b) ⊆ Ker δi.

To see the converse we recall that the construction of δi(ε) yields two ho-momorphisms, namely ` : Qi → Y such that f = b` and h : Qi+1 → X suchthat ah = `ϕi+1. Now, δi(ε) = 0 holds if and only if the factorization hbelongs to Im Hom(ϕi+1, X), that is, h = gϕi+1 for some g ∈ Hom(Qi, X).Consequently, for `′ = ag we have f = b` = b(`− `′) since b`′ = bag = 0 and(`− `′)ϕi+1 = ah− ah = 0. This shows that `− `′ ∈ Ker Hom(ϕi+1, Y ) andf ∈ Im Exti(M, b).

(iii) For each i ≥ 0, the sequence

Exti(M,Z)δi−→ Exti+1(M,X)

Exti+1(M,a)−−−−−−−−→ Exti+1(M,Y )

is exact. Let η ∈ Exti+1(M,X) be represented by h : Qi+1 → X satisfyinghϕi+2 = 0. Assume that h = δi(ε) for some ε ∈ Exti(M,Z), that is,ε is represented by a homomorphism f : Qi → Z and by construction wehave ` : Qi → Y such that f = b` and ah = `ϕi+1. The latter equationshows ah ∈ Im Hom(ϕi+1, Y ) and therefore Exti+1(M,a)(h) = 0 showingIm δi ⊆ Ker Exti+1(M,a).

For the converse inclusion we assume that η lies in Ker Exti+1(M,a), thatis, ah = `′ϕi+1 for some `′ : Qi → Y . Then set f ′ = b`′ with equivalenceclass ε′ ∈ Exti(M,X) and observe that by the construction of δi(ε

′) andLemma 5.19 we have η = δi(ε

′). This shows that shows Ker Exti+1(M,a) ⊆Im δi.

The infinite exact sequence of Theorem 5.20 is called the long exact se-quence obtained by applying Hom(M, ?) to (5.1). Taking an injectiveresolution of M we get connecting maps δ′i : ExtiA(M,X) → Exti+1

A (M,Z)and then a second long exact sequence

104


Exti−1A (X,M)

ExtiA(Z,M) ExtiA(Y,M) ExtiA(X,M)

Exti+1A (Z,M)-

- - -

-

-· · ·

· · ·

δ′i−1

δ′i

which starts with 0 → Hom(Z,M) → Hom(Y,M) → · · · , and is called thelong exact sequence obtained from (5.1) by applying Hom(?,M).

Exercises

5.4.1 Use the long exact sequences to prove that for the A-modules of a short exactsequence (5.1) the following inequalities hold:

pdimY ≤ maxpdimX,pdimZ,idimY ≤ maxidimX, idimZ.

5.4.2 Show by induction on the dimension of an A-module M that pdimM ≤maxpdimS1, . . . ,pdimSn, where S1, . . . , Sn are the simple A-modules, see Propo-sition 4.34.

105


106

Chapter 6

The Auslander-Reiten theory

There is a series of notions and results named after Auslander and Reiten, allof them constitute the Auslander-Reiten theory. They concern the structureof module categories and yield a remarkable deep insight. Since it is validin full generality it can be considered as the crown jewel of representationtheory.

6.1 The Auslander-Reiten quiver and the infiniteradical

As in the previous section, A denotes a finite-dimensional algebra over somefield K. For any isomorphism class of indecomposables in modA, fix onerepresentative and denote by indA the full subcategory of modA given bythose representatives.

For a full subcategory D ⊆ C we denote by add D the additive closure ofD , that is, the smallest full subcategory of C , which contains each object ofD and is closed under direct sums and isomorphisms: if x and y are objectsin D and x ⊕ y ∈ C a direct sum, then x ⊕ y ∈ add D ; similarly, if x ∈ Dand x′ ∈ C is isomorphic to x, then x′ ∈ add D . Note that the conditionsrefer only to objects, since only full subcategories of C are involved.

By Lemma 4.14 the category indA ⊂ modA is a small category and there-fore indA is a spectroid and although it is usually not a finite one (namelyif A is not of finite representation type), we can still associate the quiverQindA to it, which is called the Auslander-Reiten quiver of A or some-times representation quiver and denoted by ΓA.

107

6 The Auslander-Reiten theory

The vertices of the quiver ΓA are the chosen representatives of isomorphismclasses of indecomposables which constitute the objects of indA. The arrowsin indA are morphisms which lie in rad(indA) but not in rad2(indA). Wehave called such morphisms irreducible in Section 2.6. Note that a homo-morphism is radical in indA if and only if it is radical in modA. Thereforerad(indA)(M,N) = rad(modA)(M,N).

Examples 6.1 (a) The quiver given in (2.6), at the end of Section 2.6,

shows the Auslander-Reiten quiver ΓA for the path algebra A = K−→A 5 of

the linearly oriented quiver−→A 5, which is, as we know from Example 3.2 (a),

isomorphic to the algebra A of lower triangular matrices.

(b) Example 3.32 shows a quotient of K−→A 5. The Auslander-Reiten quiver

is shown in (3.3). ♦

The quiver ΓA decomposes into components in a straightforward way: twovertices X and Y lie in the same component if there is a “unoriented path”of irreducible morphisms between them, more precisely, if there are objectsX = Z0, Z1, . . . , Zt−1, Zt = Y such that for each i = 1, . . . , t there is anarrow (an irreducible morphism) Zi−1 → Zi or an arrow Zi → Zi−1. Acomponent is finite if it only contains finitely many vertices, otherwise itinfinite.

Now, ΓA encodes many information of modA, but in the case of infiniterepresentation type not all information. Remember that by Corollary 3.24the radical of indA is nilpotent if indA is finite (that is, if A is of finiterepresentation type). This no longer holds if A is of infinite representationtype. One defines the infinite radical rad∞A by

rad∞A (X,Y ) =⋂n>0

radn(indA)(X,Y ).

It contains some information not given by ΓA, namely the morphisms be-tween different components or those which start in a component and end inthe same component but still cannot be written as composition of finitelymany irreducible morphisms.

Exercises

6.1.1 Let M,N ∈ modA be two indecomposable A-modules and f : M → N aradical homomorphism. Prove the following characterization: f does not belong to

108

6.2 Example: the Kronecker algebra

rad(modA) if and only if each factorization f = hg with g : M → L and h : L→ N(where L is not necessarily indecomposable) implies that either g is a section(sections are also called split mono) (that is, g admits a left inverse) or h is aretraction (retractions are also called split epi) (that is, h admits a right inverse).

6.1.2 Calculate the Auslander-Reiten quiver of the two subspace algebra, that is,the path algebra of the quiver Q : 1 → 3 ← 2. Recall that the indecomposablerepresentations are given in Proposition 2.5. For this you have to calculate first allmorphisms between the indecomposables. Note that all calculations can be donedirectly in repQ since this category is equivalent to modA.

6.1.3 Let Q be the quiver of an algebra A. The algebra is called connected if Qis a connected. Show, that if Q is not connected then A = A1 × A2 and ΓA is thedisjoint union of ΓA1

and ΓA2.


We want to calculate the Auslander-Reiten quiver for a very interestingalgebra: the Kronecker algebra. For this, all morphism spaces are calculatedexplicitely. The calculations are rather lengthy and thus given only partiallybut with sufficient details for the reader to be able to fill in all the missingparts. If you do not want to follow all these calculations, you should at leastappreciate the taste of pure linear algebra in them and read the main resultTheorem 6.5 carefully before moving on to the next section.

The Kronecker algebra shall be denoted by A for the sake of simplicity. ByTheorem 3.3, the category modA is equivalent to repQ, where Q is theKronecker quiver, see Example 2.1 (b).

Recall also from Proposition 2.13 that repQ is equivalent to the categoryMQ which, in this case, is the Kronecker problem studied extensively inSection 1.4. The indecomposables of the Kronecker problem were classi-fied in Proposition 1.6. We translated these solutions to the language ofrepresentations, see Example 2.2 and Exercise 2.2.2.

Hence, the following representations for n ∈ N and m ∈ N>0 give a complete

109


set of pairwise non-isomorphic indecomposable representations of Q:

Rm,λ:Km Km--

1m

J(m,λ)

Rm,∞:Km Km--

J(m, 0)

1m

Qn:Kn Kn+1--

[1nz>

][z>

1n

] Jn:KnKn+1 --

[z 1n]

[1n z]

(6.1)

where z denotes a zero vector with n rows.

In order to determine ΓA we have to calculate the morphism spaces betweenall these representations. The calculations are based on induction and arerather tedious. After you have seen some of them you should be able towork out the rest by your own, which is why we present only some of them.

Lemma 6.2. If M is an indecomposable representation of Q such that thereexists a non-zero morphism f : J` → M , then M is isomorphic to Jm forsome m ≤ `. If m ≤ `, the space HomQ(J`, Jm) has dimension `−m+1 andconsists of morphisms in rad`−m. As a consequence, there are two arrowsJ` → J`−1 in the Auslander-Reiten quiver ΓA for each ` ≥ 1. Furthermore,if J` → Jm is an arrow in ΓA, then m = `− 1.

Proof. A morphism f : J` → Jm is given by two matrices A ∈ K(m+1)×(`+1)

and B ∈ Km×` such that

B · [1` z] = [1m z] ·A and (6.2)

B · [z 1`] = [z 1m] ·A (6.3)

Note that the right hand side of (6.2) (resp. of (6.3)) is obtained by elimi-nating the last (resp. first) row from A whereas the left hand side is simplyadding a zero column to B after the last column (resp. before the first col-umn). Now consider the first column of A. From equation (6.2) we get thatthe first column of A looks as shown in the following on the left, whereasequation (6.3) gives the right:

[B11 B21 · · · Bm1 ?]> = [? 0 · · · 0 0]>,

hence Bi1 = 0 for i > 1. If we look at the second column of A, we get:

[B12 B22 · · · Bm2 ?]> = [? B11 B21 · · · Bm1]> = [? B11 0 · · · 0]>,

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which implies that B22 = B11 and Bi2 = 0 for i > 2. Inductively, we getBij = Bi−1,j−1 for i ≤ j and Bij = 0 for i > j. Now, if we start comparingthe last column of A we get

[0 0 · · · 0 ?]> = [? B1` B2` · · · Bm`]>,

hence Bi` = 0 for i < m. Therefore, we also have Aij = 0 if m − i > ` − j.This shows that if m > `, then both matrices A and B are identically zero.If m ≤ `, then the matrix A is of the form β`+1(B11, B12, . . . , B1,`−m+1),where

βu(a0, a1, . . . , ar) =

a0 a1 · · · ar

a0 a1 · · · ar· · ·· · ·

· · ·a0 a1 · · · ar

∈ K(u−r)×u (6.4)

(the big circles indicate that all entries in that region are zero). Fur-thermore B = βm(B11, B12, . . . , B1,`−m+1). As a result, we obtain thatHomQ(J`, Jm) = 0 for ` < m and dimK HomQ(J`, Jm) = ` − m + 1 for` ≥ m.

We now show that each morphism f : J` → Qm is zero. If f is given by thetwo matrices A ∈ Km×(`+1) and B ∈ K(m+1)×`, then they must satisfy

B · [1` z] =

[1mz>

]·A and (6.5)

B · [z 1`] =

[z>

1m

]·A. (6.6)

If we extract the first row of (6.6), we see that on the left hand side we havea zero row, thus the first row of B is zero. Now the first row of (6.5) looksas follows

[A11 · · · A1,`+1] = [B11 · · · B1` 0] = [0 · · · 0 0]

and hence also the first row of A is zero. Now repeat the same for the secondrows. Iteratively, we find that both matrices A and B must be zero.

The proof that all morphisms f : J` → Rm,λ are zero is similar and left to thereader in Exercise 6.2.1. Therefore, the only non-zero morphisms startingin J` must end in some Jm for m ≤ `.Note also that HomQ(J`, J`) = K ·1J` . Therefore, and since HomQ(J`,M) =0 for each indecomposable representation except for M = Jm with ` ≥ m

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we have

rad2(J`, Jm) =∑

a : `>a>m

HomQ(Ja, Jm) HomQ(J`, Ja).

A direct matrix calculation shows that

βu+r(b0, . . . , bs)βu(a0, . . . , ar) = βu(c0, . . . , cr+s), (6.7)

where cn =∑n

i=0 bn−iai for n = 0, . . . , r+ s with the convention that ai = 0for i > r and similarly bj = 0 for j > s. Since morphisms between non-isomorphic indecomposables are always radical, we have that β`(1, 0) andβ`(0, 1) are radical. Moreover, it follows from (6.7), that

β`−r−s(0, 1) · · ·β`−r−1(0, 1)β`−r(1, 0) · · ·β`(1, 0) = β`(0, . . . , 0︸︷︷︸s zeros

, 1, 0, . . . , 0︸︷︷︸r zeros

).

This implies that

Hom(J`, Jm) = rad`−m(J`, Jm).

Since the space HomQ(J`, J`−1) has dimension two and rad2(J`, J`−1) = 0,we get two arrows J` → J`−1 in ΓA, namely β`(1, 0) and β`(0, 1).

The proof of the following result is completely analogous and therefore leftto the reader in Exercise 6.2.2.

Lemma 6.3. If M is an indecomposable representation of Q such that thereexists a non-zero morphism f : M → Qm then M is isomorphic to Q` forsome ` ≤ m. If ` ≤ m, the space HomQ(Q`, Qm) has dimension m −` + 1 and consists morphisms in radm−`. As a consequence, there are twoarrows Qm−1 → Qm in the Auslander-Reiten quiver ΓA for each m ≥ 1.Furthermore, if Q` → Qm is an arrow in ΓA, then ` = m− 1.

The next is the third and last auxiliary result needed for the full determi-nation of ΓA.

Lemma 6.4. If f : R`,λ → Rm,µ is a non-zero morphism for λ, µ ∈ K∪∞,then λ = µ. If f is irreducible, then m = ` ± 1. Moreover, for each ` ≥ 1there exists precisely one arrow R`,λ → R`+1,λ and one arrow R`+1,λ → R`,λ.

Proof. Assume first that λ, µ 6= ∞. A morphism f : R`,λ → Rm,µ consistsof two matrices A,B ∈ Km×` satisfying two conditions coming from the

112


commutative diagrams (2.4) for the two arrows. The first condition is A =B, so the second is

AJ(`, λ) = J(m,µ)A. (6.8)

Now assume additionally that λ 6= µ. Then the last entry in the first row of(6.8) is λAm1 = µAm1, hence Am1 = 0. Now look at the entry in position(m − 1, 1). We find λAm−1,1 + Am1 = µAm−1,1 and hence Am−1,1 = 0.Proceeding this way we obtain Ai1 = 0 for all i. Observe that therefore(J(m,µ)A)i2 = µAi2 for all i and we may repeat our procedure for thesecond column showing Ai2 = 0 for all i. Inductively, we get that A is thezero matrix.

If λ =∞ 6= µ or λ 6=∞ = µ a very similar argument shows the same result.Hence

Hom(R`,λ, Rm,µ) = 0, if λ 6= µ.

Now, if λ = µ 6=∞, then each morphism f : R`,λ → Rm,λ is given by a pairof matrices (A,A) which satisfies

A(λ1` + J(`, 0)) = (λ1m + J(m, 0))A.

This is equivalent to AJ(`, 0) = J(m, 0)A. Proceeding from the lower leftcorner up and to the right we conclude that Aij = 0 if one of the fol-lowing conditions is satisfied: i > j or m − i < ` − j. For all other en-tries Aij = Ai+1,j+1 holds. Hence, if ` ≤ m, the matrix A is of the formA = γm,`(A11, . . . , A1`) where

γm,`(a0, . . . , a`−1) =

a0

a0

a1

a1

· · ·· · · · ·

a`−1· · ·

···

∈ Km×`

and if ` > m, then A is of the form A = γm,`(Am`, Am−1,`, . . . , A1`) wherein this case the matrix γm,`(a0, . . . , am−1) is defined as

γm,`(a0, . . . , am−1) =

a0

a0

a1

a1

· · ·· · · · ·

am−1· · ·

···

∈ Km×`.

113


A similar calculation deals with the case when λ = µ = ∞. We concludethat Hom(R`,λ, Rm,λ) has dimension min(`,m).

A note of caution: As we see, the same pair (A,A) of matrices constitutes amorphism R`,λ → Rm,λ, regardless of the value λ ∈ K ∪ ∞. However, asmorphisms they are different, as a morphism always “knows” to which pairof objects it belongs.

Now, observe that in the sum on the right hand side of

rad(R`,λ, Rm,λ) =∑

M∈indQ

rad(M,Rm,λ) rad(R`,λ,M)

most of the summands are zero, since by Lemma 6.2 we haveHom(Jn, Rm,λ) = 0 for all n, by Lemma 6.3, Hom(R`,λ, Qn) = 0 for alln and by the above Hom(R`,λ, Rn,ν) = 0 for all ν 6= λ. Hence only thesummands for M = Rn,λ may contribute.

Similarly, as in the proof of Lemma 6.2, we see that the morphisms given bythe matrices γm,`(a0, . . . , ar) behave well under composition and conclude

that Hom(R`,λ, Rm,λ) = rad|`−m|(R`,λ, Rm,λ) and rad|`−m|+1(R`,λ, Rm,λ) =0. Hence, there is exactly one arrow R`,λ → R`+1,λ and one R`+1,λ → R`,λand no other arrow between two vertices of the form R`,λ and Rm,µ.

Theorem 6.5. The Auslander-Reiten quiver ΓA of the Kronecker algebraA looks as follows:

P

Q0

Q1

Q2

Q3

Q4

JJJJ JJJJ

······

I

J0

J1

J2

J3

J4

J

JJJ

J

JJJ

··· ···

Rλ (λ ∈ K ∪ ∞)R1,λ

R2,λ

R3,λ

R4,λ

6?

6?

6?

?······

More precisely, there are two special components: the preprojective com-ponent P which contains all A-modules Qn and two arrows Qn → Qn+1

for n ≥ 0, and the preinjective component I which contains all Jn andtwo arrows Jn+1 → Jn for n ≥ 0. Moreover, there is a family (Rλ)λ∈K∪∞

114


of regular components, each Rλ contains all Rm,λ and two arrows be-tween Rm,λ and Rm+1,λ, one in each direction for m ≥ 1. And these are allarrows in ΓA.

All morphisms which go in the picture from right to left are zero and eachmorphism Q` → Jn factors, for each λ, through some Rm,λ. Furthermore,for λ 6= µ all morphisms R`,λ → Rm,µ are zero.

Proof. We have already seen in Lemma 6.2, 6.3 and 6.4 that in ΓA, thereare arrows as described in the statement and that the morphisms from rightto left and between different Rλ and Rµ are zero.

It remains to see that there are no further arrows and to prove the statedfactorization property. For this, we first show these properties for the caseλ = 0, which makes calculations easy. A morphism f : Q` → Rm,0 is givenby two matrices A ∈ Km×` and B ∈ Km×(`+1) which satisfy

A = B

[1`z>

],

J(m, 0)A = B

[z>

1`

].

We observe that the first equation states that A equals the first ` columnsof B. Substitution of the first equation in the second and an inductiveargument yield that B satisfies Bij = 0 if i + j ≥ m and Bij = Bi−1,j+1

for all 1 < i ≤ m, 1 ≤ j ≤ `. Typical examples of such matrices are

σ(i)m,`+1 ∈ K

m×(`+1) with entries given by

(σ

(i)m,`+1

)ab

=

1, if a+ b− 1 = i,

0, else.(6.9)

The pairs

s(i)m,` =

(σ

(i)m,`, σ

(i)m,`+1

), for i = 1, . . . ,m (6.10)

give a basis of HomQ(Q`, Rm,0).

Furthermore, a direct calculation shows that

σ(i)m,` = γm,m+1(1, 0, . . . , 0)σm+1,`,

and therefore s(i)m,` belongs to rad2. Hence none of the morphisms s

(i)m,` can

be irreducible and consequently, in ΓA, there is no arrow from Q` to Rm,0for any `,m.

115


Now, a completely analogous calculation shows that in ΓA there is no arrowfrom Rm,0 to Jn for any m,n. Moreover, if (C,D) : Rm,0 → Jn is a mor-phism, then C satisfies Cij = 0 for i > j and Cij = Ci+1,j+1 for 1 ≤ i < n,

1 ≤ j ≤ m. Typical examples of such matrices are ρ(j)n+1,m ∈ K(n+1)×m for

j = 1, . . . ,m whose entries are given by

(ρ

(j)n+1,m

)cd

=

1, if d− c+ 1 = j,

0, else.(6.11)

Furthermore, we define the morphisms r(j)m,n = (ρ

(j)n+1,m, ρ

(j)n,m) for j =

1, . . . ,m. Note that they form a basis of Hom(Rm,0, Jn).

Finally, we determine all morphisms (E,F ) : Q` → Jn. With similar argu-ments as seen before, we conclude that the matrix E satisfies Eij = Ei−1,j+1

(for all i and j, where this makes sense). Typical such matrices are σ(i)n+1,`

given as in (6.9), with the difference that all indices i = 1, . . . , ` + n arepossible. The pairs

t(i)n,` =

(σ

(i)n+1,`, σ

(i)n,`+1

), for i = 1, . . . , `+ n (6.12)

form a basis of Hom(Q`, Jn).

Now it is easy to see that for m = ` + n, we have ρ(1)n+1,mσ

(i)m,` = σ

(i)n+1,` and

ρ(1)n,m+1σ

(i)m+1,`+1 = σ

(i)n,`+1 and therefore

t(i)n,` = r(1)

n,ms(i)m,`, for i = 1, . . . ,m = `+ n (6.13)

This shows that each morphism Q` → Jn factors through R`+n,0, that is thefactorization property for λ = 0.

To obtain the analogous statements for all other possible values of λ we use atrick. Recall that A is the path algebra A = KQ, where Q is the Kroneckerquiver:

Q : 1 2 .--α

β

Consider the K-linear map Hλ : A → A defined by Hλ(ei) = ei for i =1, 2, Hλ(α) = α and Hλ(β) = β + λα. Then it follows that Hλ(δγ) =Hλ(δ)Hλ(γ) for any two paths δ, γ and hence for any two elements of thepath algebra. Consequently Hλ is a homomorphism of K-algebras. SinceHλ has an inverse, namely H−λ, it is an isomorphism of K-algebras.

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6.3 The Auslander-Reiten translate

Now, ifM is anA-module with structure mapA×M →M, (a,m) 7→ a·m, wecan define a second structure map A×M →M, (a,m) 7→ a∗λm := Hλ(a)·mand shall denote by Mλ the abelian group M together with this map. Iff : M → N is a homomorphism of A-modules, then also fλ = f : Mλ → Nλ,since

a ∗λ f(m) = Hλ(a) · f(m) = f(Hλ(a) ·m) = f(a ∗λ m).

Hence, we get a functor ?λ : modA → modA which has ?−λ as inverse.These functors are therefore isomorphisms, in particular they send inde-composables to indecomposables. Hence (Q`)

λ is again indecomposable andits vector spaces have dimensions ` and `+ 1 in the vertices 1 and 2 respec-tively. Since there is only one such indecomposable in indA, we must have(Q`)

λ ' Q`. Similarly (Jn)λ ' Jn. We now describe V = (Rm,0)λ. It isclear that we only have to determine the multiplication with β, and if (v1, v2)denotes a general element in V , then only β ∗λ v1 has to be considered:

β ∗λ v1 = Hλ(β) · v1 = (β + λα) · v1 = J(m,λ)v1,

that is, (Rm,0)λ = Rm,λ. Since ?λ is a functor, we get the statements for anyλ 6= ∞. For this last case the interchange automorphism I : A → A, whichinterchanges α and β may be considered covering this last case in the sameway.

Remark 6.6. We have presented a “low-tech” proof. Other approachescan for example be found in Ringel’s book, [22], where much more theoryis developed before and then used in the proof. We took this rather bumpyride to show that the representation’s point of view reduces many questionseffectively to linear algebra, where they might be solved. In the following,we present gradually more theory, as it offers more insight, and often comeback to this particular example. ♦

Exercises

6.2.1 Verify that all morphisms f : J` → Rm,λ are zero.

6.2.2 Prove Lemma 6.3.


As usual, let A be a finite-dimensional K-algebra. A sequence of morphismsin modA

M0f1−→M1

f2−→M2 → . . .→Mt−1ft−→Mt (6.14)

117


is called exact if Ker fi = Im fi−1 holds for all i = 1, . . . t.

Example 6.7. The sequence

0→MidM−−→M → 0

is always exact for each module M ∈ modA. Also, for each morphismf : M → N the sequence

0→ Ker fι−→M

f−→ Nπ−→ Coker f → 0,

where ι and π denote the canonical inclusion and projection respectively, isan exact sequence. ♦

By Lemma 4.9 a sequence 0→Mf−→ N (resp. M

f−→ N → 0) is exact if andonly if f is injective (resp. surjective).

A projective presentation of a module M ∈ modA is an exact sequence

Qµ−→ P

ε−→M → 0 (6.15)

where P and Q are projectives. Such a projective presentation is minimalif ε is a projective cover of M and µ is a projective cover of the object Ker ε.This shows that a projective presentation always exists. The dual conceptis called injective presentation.

Example 6.8. We continue Example 4.31, where we calculated the projec-tive cover of M = [2, 3] to be ε = ι2,32,5 : [2, 5]→ [2, 3]. Its kernel is [4, 5] whichis projective. Hence

[4, 5]→ [2, 5]→ [2, 3]

is a minimal projective presentation of [2, 3]. ♦

Note that in a projective presentation (6.15) it is enough to give the mor-phism µ, since ε = Coker(µ) is determined by µ up to isomorphism. Now,if ϕ1 : Q1 → P1 is a projective presentation of M1 and ϕ2 : Q2 → P2 a pro-jective presentation of M2 then clearly ϕ1 ⊕ ϕ2 : Q1 ⊕ Q2 → P1 ⊕ P2 is aprojective presentation of M1 ⊕M2. Conversely, if a projective presenta-tion ϕ admits a decomposition ϕ = ϕ1 ⊕ ϕ2, then the presented moduledecomposes.

We are now ready to describe an important construction for the study of themodule category modA, namely that of the Auslander-Reiten translate,

118


which will be denoted by τA: given an module M ∈ modA, we first take aminimal projective presentation (6.15), apply the Nakayama functor νA tothe morphism µ : Q→ P and then take the kernel

NKer(νAµ)−−−−−−→ νAQ

νAµ−−→ νAP. (6.16)

We then define the Auslander-Reiten translate τAM = N of M , whereN is the kernel in (6.16). The important properties of the Auslander-Reitentranslate, which is also called AR-translate for short, are gathered in thefollowing result.

Proposition 6.9. Let A be a finite-dimensional K-algebra. The AR-translate τAM of M ∈ modA is independent of the choice of the mini-mal projective presentation of M . We have τAM = 0 if and only if M isprojective. Furthermore, the AR-translate induces a bijection between theindecomposable non-projectives and the indecomposable non-injectives, upto isomorphism.

Proof. The independence of the choice of projective presentation followsfrom Corollary 4.29. Now, if M is projective then idM : M →M is a projec-tive cover and 0 → M → M a projective presentation of M . ConsequentlyτAM = 0. Conversely, if Ker νAµ = 0, then νAµ is injective and thereforesplits by the dual of Exercise 4.5.1, that is,

νAµ '[idνAQ

0

]: νAQ→ νAQ⊕ νAQ′ = νAP.

Consequently ε = Cokerµ = [0 idQ′ ] : Q ⊕Q′ → Q′ = M and M is projec-tive.

Now, suppose that M is an indecomposable non-projective A-module and(6.15) a minimal projective presentation of M . Then µ cannot decomposesince otherwise M decomposes. Therefore νAµ does not decompose showingthat τAM is indecomposable or zero. However, the latter is not possiblesince M was assumed to be non-projective.

Using dual arguments, we can start with any A-module L and construct theminimal injective presentation

0→ Lϕ−→ I

λ−→ J,

then apply the inverse of the Nakajama functor ν−1A and construct τ−AL =

Coker(ν−1A λ). Dually, we have τ−AL = 0 if and only if L is injective.

119


Hence for M an indecomposable non-projective τAM is indecomposable andnon-injective. Consequently, τA gives a bijection between the isomorphismclasses of indecomposable non-projectives and the isomorphism classes ofindecoposables non-injectives.

Remark 6.10. The notation τ−A might seem strange at first sight, butit remains us of the fact that τ−A is not really an inverse of τA: for eachprojective A-module M we have τAM = 0 and there exists no A-module Nsuch that τ−N = M . Therefore τA and τ−A are not properly inverse to eachother. However, we will use τ−` instead of the clumsy (τ−)`. ♦

Example 6.11. As in the Examples 4.31 and 6.8 let A = K−→A 5 and

M = [2, 3]. Then ι : P4 = [4, 5] → [2, 5] = P2 is the minimal projectivepresentation. Hence we obtain νAι : I4 = [1, 4] → [1, 2] = I2 by applyingthe Nakayama functor and τAM = Ker νAι = [3, 4]. We can do this with allindecomposables and obtain the following picture:

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R[5, 5]

[4, 5]

[3, 5]

[2, 5]

[1, 5]

[4, 4]

[3, 4]

[2, 4]

[1, 4]

[3, 3]

[2, 3]

[1, 3]

[2, 2]

[1, 2]

[1, 1]τA τA τA τA

τA τA τA

τA τA

τA

(6.17)

The Auslander-Reiten quiver should always be considered as equipped withthe AR-translate. But there is still more to explore and this will be done inthe next chapter. ♦

Exercises

6.3.1 Show that a sequence Uf−→ V

g−→W of representations of a quiver Q is exact

if and only if for each vertex i the sequence Uifi−→ Vi

gi−→Wi is exact.

6.3.2 Calculate the Auslander-Reiten translate of all indecomposable non-projec-tive A-modules, where A is the algebra considered in Example 3.32.

6.3.3 Calculate the Auslander-Reiten translate of the indecomposable non-projec-tives over the two subspace algebra.

6.3.4 Show that if in the exact sequence (6.14) the two terms Mi−1 and Mi+1 arezero, then also Mi.

120

6.4 Auslander-Reiten sequences


Let A be a finite-dimensionalK-algebra. Note that for each A-moduleM thespace HomA(M,A) is a right A-module, where the right multiplication frfor f ∈ HomA(M,A) and r ∈ A is given as fr : M → A, s 7→ f(s)r. In otherwords, HomA(M,A) is a left Aop-module and therefore D HomA(M,A) isa left A-module. The following result shows that D HomA(?, A) generalizesthe Nakayama functor νA defined in Section 4.5.

Lemma 6.12. Let A be a finite-dimensional K-algebra. For each projec-tive A-module P we have νAP ' D HomA(P,A) as A-modules and for eachhomomorphism ϕ : P → P ′ we have νAϕ ' D HomA(ϕ,A).

Proof. For the indecomposable projective Px associated to the vertex x weget by Yoneda’s Lemma 4.24

HomA(Px, A)Φx,A−−−→ Ax = exA, f 7→ f(ex).

Since Φx,A(fr) = fr(ex) = f(ex)r = Φx,A(f)r for each r ∈ A this bijectionpreserves the right multiplication. Dualization yields an isomorphism of leftA-modues

νAPx = Ix = DexADΦx,A−−−−→ D HomA(Px, A).

This isomorphism is functorial in x, that is, if α : x → y is an arrow of thequiver of A then Pα : Py → Px is a homomorphism and νAPα = Iα = D(α·?)corresponds to D HomA(Pα, A), see Remark 4.25 (b). The result follows now,as each projective A-module is decomposed into a direct sum of the Px.

We can now define the Nakayama functor for the whole module categoryνA : modA→ modA by

νA = D HomA(?, A). (6.18)

We give one nice property of the Nakayama functor at once.

Lemma 6.13. For each exact sequence Xa−→ Y

b−→ Z → 0 the Nakayamafunctor yields an exact sequence

νAXνAa−−→ νAY

νAb−−→ νAZ → 0.

121


Proof. We know from Theorem 5.20 that

0→ Hom(Z,A)Hom(b,A)−−−−−−→ Hom(Y,A)

Hom(a,A)−−−−−−→ Hom(X,A)

is exact. Applying the dualization over the ground field reverses the arrowsand preserves exactness, see Exercise 5.1.5, hence the result.

With the new definition of the Nakayama functor (6.18) we can also gener-alize Lemma 4.24. For this let L,M be A-modules. Note that any homo-morphism A → L is given as mul` : r 7→ r` for some ` ∈ L. Hence for agiven homomorphism g : M → A and a given element ` ∈ L, we can definethe homomorphism

mul` g : M → L,m 7→ g(m)`. (6.19)

Proposition 6.14. For any two A-modules L and M there exists a naturalmap

αL,M : D HomA(M,L)→ HomA(L, νAM), (6.20)

given by αL,M (ϕ) = ϕ(mul`?). This map is functorial and additive in botharguments and it is a bijection if M is projective. Furthermore, the kernelof αL,M can be described explicitely

KerαL,M = ϕ | ϕ(projA(M,L)) = 0,

where projA(M,L) is the subset of HomA(M,L) given by those homomor-phisms which factor over some projective A-module.

Proof. We start by looking that the explicit description makes sense. Fora given ϕ ∈ D HomA(M,L) the value αL,M (ϕ) has to be a homomorphismL→ D HomA(M,A), that is, for each ` ∈ L a linear map HomA(M,A)→ Khas to be given. This is done by g 7→ ϕ(mul` g).

Now, we look at the case where M is a projective indecomposable M = Px =Aex. By Yoneda’s Lemma 4.24 we have the following sequence of functorialK-linear bijections

D HomA(Px, L)F=DΦ−1

x,L−−−−−−→ DLxG=Ψ−1

x,L−−−−−→ HomA(L,DexA). (6.21)

Using Lemma 6.12 we have a bijection DΦx,A : DexA→ D HomA(Px, A) andhence can prolongate the sequence (6.21) one step further:

HomA(L,DexA)H=HomA(L,DΦx,A)−−−−−−−−−−−−−→ HomA(L,D Hom(Px, A)). (6.22)

122


Therefore the composition HGF is K-linear, bijective and functorial. Wenow show that HGF is in fact αL,Px .

The inverse of Φx,L was described in the proof of Yoneda’s Lemma 4.24 andtherefore for ϕ ∈ HomA(Px, L) we have

F (ϕ) : Lx → K, ` 7→ ϕ(mul`),

see the following picture.

HomA(Aex, L)

K

Lx(r 7→ r`)

ϕ(r 7→ r`)

`

?

?

For ρ ∈ DLx we first observe that

HG(ρ) = DΦx,A G(ρ) : L→ D Hom(Aex, A)).

The functor G = Ψ−1x,L was described in the proof of Yoneda’s Lemma 4.24.

The evaluation G(ρ) for ρ ∈ DLx is given as shown in the next picture onthe left-hand side. The right-hand side is obtained by evaluating the effectof DΦx,A.

L DexA D Hom(Aex, A)- -G(ρ) DΦx,A

`

(exA −→ Ks −→ ρ(s`)

) (Hom(Aex, A) −→ K

f −→ ρ(f(ex)`)

)- -

Now, if ρ = ϕ(r 7→ r`), then

ρ(f(ex)`) = ϕ(r 7→ rf(ex)`),

but since r = rex we have rf(ex)` = f(r)`. This shows that HGF (ϕ) is themap L→ D Hom(Aex, A), ` 7→ ϕ(r 7→ f(r)`) exactly as does αL,Px(ϕ).

For a general projective A-module the result follows now immediately sinceHGF is functorial and therefore preserves direct sums. It remains to seethat for any M the map αL,M has the kernel as indicated in the statement.

123


Thus, assume that αL,M (ϕ) = 0. This means that for all ` ∈ L and allg : M → A the value ϕ(mul` g) equals zero. Now, let f : M → L be anyhomomorphism which factors over some projective A-module Q, say f =f ′′f ′ for some f ′ : M → Q and some f ′′ : Q → L. Take any surjectivehomomorphism h = [h1 · · · hm] : Am → L and use the projectivity of Q tofactor f ′′ through h, that is, f ′′ = hk for some k = [k1 · · · km]> : Q→ Am.Then f = hkf ′ =

∑mi=1 hi(kif

′) is the sum of homomorphisms which factorover A. As we have observed before hi = mulì for some ì ∈ L. Since ϕ isadditive we see that indeed ϕ(f) =

∑mi=1 ϕ(mulì kif ′) = 0 and therefore

ϕ(proj(M,L)) = 0. The converse follows at once, since mul` g factors overthe projective A and therefore αL,M (ϕ) = 0.

Remark 6.15. For two A-modules X and Y we denote by projA(X,Y )the subspace of HomA(X,Y ) given by all homomorphisms which factorover a projective A-module. It is straightforward that projA forms anideal of the category modA. The respective quotients HomA(X,Y ) =HomA(X,Y )/ projA(X,Y ) form a category which is denoted by modA.

Similarly, injA is the ideal of all homomorphisms which factor over an in-jective module and the quotients HomA(X,Y ) = HomA(X,Y )/ injA(X,Y ).form a category which is denoted by modA. Both quotients, modA andmodA, are called stable module categories.

This allows to describe the kernel of αL,M in the following way

KerαL,M = D HomA(M,L).

♦

We have now gathered sufficient observations from homological algebra to setout and prove the most important result on the structure of the Auslander-Reiten quiver of A.

An exact sequence 0 → La−→ E

b−→ M → 0 is called a Auslander-Reitensequence if it is a short exact non-split sequence with indecomposable endterms and if any radical homomorphism a′ : L → L′ factors through a andany radical homomorphism b′ : M ′ →M factors through b. Note that sinceL and M are indecomposable a homomorphism a′ : L→ L′ is radical if andonly if it is not a section, see Exercise 6.1.1 and similarly b′ : M ′ → M isradical if and only if b′ is not a retraction. The following result not onlyasserts the existence of such sequences, but it also implies that the end termsof such a sequence are always linked by the Auslander-Reiten translate.

124


Theorem 6.16 (Auslander-Reiten). Let A be a finite-dimensional algebraover K.

(a) For any two modules L,M ∈ modA, there are bijections

Ext1A(L, τAM) −→ DHomA(M,L),

Ext1A(τ−AL,M) −→ DHomA(M,L)

which are functorial in both arguments.

(b) For any non-projective indecomposable module M there exists an Aus-lander-Reiten sequence 0→ τAM → E →M → 0.

(c) For any non-injective indecomposable module L there exists an Aus-lander-Reiten sequence 0→ L→ E → τ−AL→ 0.

Proof. Clearly, it is enough to prove (a) for indecomposable M because bothsides preserve direct sums. If M is in addition projective, we trivially haveequality since both sides are zero. So, suppose that M is non-projective andindecomposable.

Recall from Section 6.3 the construction of τAM : We have to take a minimalprojective presentation Q

µ−→ Pε−→M → 0 of M . Using Lemma 6.13 we get

an exact sequence

0→ τAM → νAQνAµ−−→ νAP

νAε−−→ νAM → 0 (6.23)

since by definition τAM = Ker νAµ. If we compose νAε : νAP → νAM withthe injective hull ι : νAM → J of νAM , we get the starting of an injectiveresolution of τAM and hence can calculate by (5.12)

Ext1A(L, τAM) = Ker HomA(L, ι νAε)/ Im HomA(L, νAµ). (6.24)

Note that Ker HomA(L, ι νAε) = Ker HomA(L, νAε) since ι is injective. Itfollows from the functoriality of αL,? that the following diagram is commu-tative.

HomA(L, νAQ) HomA(L, νAP ) HomA(L, νAM)

D HomA(Q,L) D HomA(P,L) D HomA(M,L) 0- -

- -

-

? ? ?

αL,Q αL,P αL,M

D HomA(µ,L) D HomA(ε, L)

HomA(L, νAµ) HomA(L, νAε)

125


Furthermore, the rows are exact. In the following sequence of K-linearbijections, the first is due to the bijectivity of αL,Q and αL,P , the secondand third to the exactness of the upper row in the diagram.

HomA(L, νAP )/ Im HomA(L, νAµ) ' D HomA(P,L)/ Im D Hom(µ,L)

' D HomA(P,L)/Ker D Hom(ε, L)

' D HomA(M,L).

The following sequence of K-linear bijections yields the first formula of (a).The first bijection is due to (6.24), the last to Remark 6.15.

Ext1A(L, τAM) = Ker Hom(L, νAε)/ Im Hom(L, νAµ)

' Ker (αL,M D Hom(ε, L)) /Ker D Hom(ε, L)

' KerαL,M

= D Hom(M,L)

Observe that all bijections are functorial in both arguments. The secondstatement is completely dual to the first. This shows part (a).

For part (b), we choose L = M . Since M is not projective the identityhomomorphism idM does not factor over a projective A-module and sinceM is indecomposable HomA(M,M) is local with maximal ideal radA(M,M).We therefore can define a non-zero linear map Ω: HomA(M,M) → K byΩ(idM ) = 1 and Ω(r) = 0 for each radical morphism r ∈ radA(M,M).

According to part (a) the linear map Ω corresponds to a short exact sequence

εM : 0→ τAMa−→ E

b−→M → 0 (6.25)

which has indecomposable end terms and does not split since Ω 6= 0. Wewill show that this is indeed an Auslander-Reiten sequence. Let g : L→Mbe a non-retraction. If we decompose L into indecomposables L =

⊕ti=1 Li

and write g = [g1 . . . gt], we have that each gi is a radical homomorphism.Therefore, if we consider

DHomA(M, gi) : DHomA(M,M)→ DHomA(M,Li),

then Ω ∈ DHom(M,M) is mapped to Ω (gi?). Certainly Ω (gi f) = 0holds for each endomorphism f of M since the composition gif is alwaysradical. This shows that DHomA(M, g)(Ω) = 0. Exploiting the functorialityof the formula given in part (a) we get that εMg = Ext1

A(g, τAM)(εM ) = 0,that is, the sequence obtained from (6.25) by taking the pull-back along g

126


and b splits. Thus by Exercise 5.2.3 the homomorphism g factors throughb. The rest, namely that every non-section τAM → L factors over a, followsdually. Part (c) is shown analogously to (b).

Remark 6.17. The Auslander-Reiten Theorem is one of the most importantresults in representation theory of finite-dimensional algebras. It is valid inbroader generality, see [2]. Note that the Auslander-Reiten formulas, aspart (a) is often called, yields a forth and fifth way to calculate extensiongroups. We will see in the next section that the Auslander-Reiten sequencesmay be used to effectively calculate parts of the AuslanderReiten quiver, atechnique which is known as knitting. ♦

Exercises

6.4.1 Let C and D are abelian categories. A functor F : C → D is called exactif each short exact sequence is mapped to a sequence which is again short exact.Show that the dualization D: vec→ vec is an exact contravariant functor.

6.4.2 Show that in the quotient category modA each projective A-module is a zeroobject.

6.4.3 Let ε ∈ Ext1A(M, τAM) be the class of the Auslander-Reiten sequence ending

in M . Show that for each radical homomorphism f : τAM → X (resp. for eachradical homomorphism g : Y → M) the sequence fε (resp. εg) splits. This is whyAR-sequences are also called almost split sequences.

127


128

Chapter 7

Knitting

The Auslander-Reiten Theorem can be translated into a combinatorial tech-nique called knitting. It yields in many concrete cases large parts or all ofthe Auslander-Reiten quiver including the structure of the contained inde-composable modules.

7.1 Source and sink maps

In this chapter we first develop a technique from Auslander-Reiten’s The-orem 6.16 which is of great importance because it allows to compute theAuslander-Reiten quiver ΓA of a finite dimensional K-algebra A in manyexamples. Later in this chapter other combinatorial concepts are introducedwhich are useful in the classification of indecomposable A-modules.

We assume that all indecomposables appearing in the considerations arevertices of ΓA. This has the advantage that two indecomposables which areisomorphic are equal, thereby simplifying the arguments. Recall from Ex-ample 3.23 that for non-isomorphic, indecomposable A-modules M and Nwe have radA(M,N) = HomA(M,N). Moreover HomA(M,M) is a finite-dimensional, local algebra with radical radA(M,M) consisting of all homo-morphisms which are nilpotent, or equivalently, which are not isomorphisms;see Proposition 3.13.

As we will see the two maps occurring in an Auslander-Reiten sequencecan be characterised in categorical terms. To do so we introduce some newterminology. Let M be an indecomposable A-module. A homomorphisma : M → N is a source map for M if it is radical, every radical homo-

129

7 Knitting

morphism g : M → X factors through a (that is, there exists f : N → Xsuch that fa = g) and every ϕ : N → N with ϕa = a is an isomorphism.The dual concept is called sink map. The source and sink maps for anindecomposable A-module are unique, see Exercise 7.1.1.

We first show that the third condition for a source or a sink map can bereplaced.

Lemma 7.1. Let M be an indecomposable A-module and N =⊕t

i=1Ni anA-module decomposed into indecomposables. Then a homomorphism f =[f1 · · · ft

]>: M → N is a source map if and only if the following three

conditions hold: (i) f is radical, (ii) every radical homomorphism g : M →N ′ factors through f and (iii) for each Y ∈ ΓA the set fi | Ni = Y ismapped to a basis of the space radA(M,Y )/ rad2

A(M,Y ) under the canonicalprojection.

Dually a homomorphism f : N → M is a sink map if and only if (i) f isradical, (ii) every radical homomorphism g : N ′ →M factors through f and(iii) for every Y ∈ ΓA the set fi | Ni = Y yields a basis of the spaceradA(Y,M)/ rad2

A(Y,M).

Proof. Assume first that the homomorphism f satisfies (i), (ii) and (iii).Further let ϕ : N → N be a homomorphism such that ϕf = f . We write ϕas matrix with entries ϕji : Ni → Nj . Now, we fix Y ∈ ΓA and observe thatwe may assume without loss of generality that Ni = Y for 1 ≤ i ≤ s andNi 6= Y for s < i ≤ t. Then for i = 1, . . . , s we have modulo rad2

A that

fi =

t∑j=1

ϕijfj ≡s∑j=1

ϕijfj

holds because ϕij is radical for all i ≤ s and j > s. For each i, j = 1, . . . , swe have that ϕij = λij idY +ρij , where λij ∈ K and ρij is radical. Thus itfollows that modulo rad2

A we have

fi ≡s∑j=1

λijfj .

These equations indicate a base change in the spaceradA(M,Y )/ rad2

A(M,Y ) showing that the matrix λ = (λij)si,j=1 ∈ Ks×s

is invertible. Therefore (ϕ)si,j=1 is also invertible and by varying Y we seethat ϕ itself is an isomorphism.

130

7.1 Source and sink maps

Now assume that f satisfies the first two conditions of a source map and thatϕf = f implies that ϕ is an isomorphism. We fix Y ∈ ΓA and assume againthat Ni = Y for 1 ≤ i ≤ s and Ni 6= Y for s < i ≤ t. To prove property(iii) assume that λi ∈ K are scalars for i = 1, . . . s such that ψ =

∑si=1 λifi

belongs to rad2A(M,Y ). We have to show that λi = 0 for all i = 1, . . . , s.

So assume otherwise: λj 6= 0 for some j ≤ s. Since ψ ∈ rad2A(M,Y ) there

are two radical homomorphisms g : M → X and h : X → Y with ψ = hg.By property (ii) there exists a factorization g′ : N → X such that g′f = g.Consequently

ψ = Λf = hg′f : M → Y,

where Λ:⊕t

i=1Ni → Y is given by Λi = λi idY for i ≤ s and Λi = 0 fori > s. Now let ιj : Y → N be the inclusion into the j-th summand Nj = Y .Then

f = (idN −ιj1

λj(Λ− hg′))f

and therefore ϕ = idN −ιj 1λj

(Λ−hg′) is an isomorphism. But an inspection

of the j-th column of (ϕih)ti,h=1 will show the contrary: clearly ϕij = 0

holds for all i 6= j and ϕjj = idY − 1λj

(λj idY −hg′j) = 1λjhg′j is radical.

Thus we have reached a contradiction and shown that f1, . . . , fs are linearlyindependent modulo rad2

A(M,Y ).

This concludes the proof of the statement for source maps and the statementfor sink maps is obtaind by dualization.

Proposition 7.2. For every indecomposable A-module a source map anda sink map exists. More precisely: if M is indecomposable non-projective,

then the homomorphism b in the Auslander-Reiten sequence τAM → Eb−→ M

is a sink map for M . If M is indecomposable projective, then the canonicalinclusion radM →M is a sink map for M . Dually, if M is indecomposable

non-injective, then the map b in the Auslander-Reiten sequence Mb−→ F →

τ−AM is a source map for M . If M is indecomposable injective, then thecanonical projection M →M/ socM is a source map for M .

Proof. Let M be an indecomposable A-module. If M is non-projective, thenby the Auslander-Reiten Theorem 6.16 there exists an AR-sequence endingin M . If we decompose the middle term into indecomposables E =

⊕ti=1Ei

this sequence looks as follows:

0→ τAMa=[a1 ··· at]>−−−−−−−−−→

t⊕i=1

Eib=[b1 ··· bt]−−−−−−−→M → 0 (7.1)

131

7 Knitting

First note that the homomorphisms bi are radical since otherwise theywould be invertible and the Auslander-Reiten sequence (7.1) would splitin contradiction to the definition. Furthermore, by the definition of AR-sequences each radical homomorphism g : X → M factors through b whichshows the second property. Now, if ϕ : E → E satisfies bϕ = b, then byLemma 3.12 there exists some integer n such that E = Im(ϕn) ⊕ Ker(ϕn).Since b(idE −ϕn) = 0 and a is a kernel of b, there exists a homomorphismc : E → τAM such that

ac = idE −ϕn.

If we restrict this equality to Kerϕn, then on the right we get a sectionwhereas on the left we get a radical homomorphism. This can only happenif this restriction is zero. Consequently Kerϕn = 0. This shows that ϕn andtherefore also ϕ itself is injective and hence bijective. It follows that themap b of the AR-sequence (7.1) is a sink map.

Now we assume that M is indecomposable and projective, say M = Px.We have to show that the inclusion ι : radPx → Px is a sink map. LetradPx =

⊕j Nj be a decomposition into indecomposable summands. Then

ι induces injective maps Nj → Px which never can be an isomorphism sincedimK Nj ≤ dimK radPx < dimK Px. Therefore ι is radical. Let π : Px → Sxbe the canonical projection and observe that if some homomorphism g : Y →Px satisfies πg 6= 0, then there exists some y ∈ Y such that g(y) = ex− r forsome r ∈ radPx = radAex ⊂ Aex. Thus g(y+ry) = ex−r+rg(y) = ex−r2,where we used r = rex. Inductively we get g(y+ry+ . . .+rn−1y) = ex−rn.Observe that rn belongs to radn Px. Since radn Px = 0 for large n we get thatthere exists an element y′ ∈ Y such that ex = g(y′). Consequently for alla ∈ Px we have a = aex = ag(y′) = g(ay′) showing that g is surjective andhence splits by Exercise 4.5.1. This shows that every radical homomorphismg : Y → Px factors over ι. Finally, if f : radPx → radPx satisfies ιf = ι,then f = idradPx since ι is injective.

The existence of source maps and the statement about their shape followsby dualization.

Remark 7.3. It follows from the previous description of source and sinkmaps that in the Auslander-Reiten quiver ΓA of a finite-dimensional K-algebra there exist only finitely many arrows with a fixed start vertex ora fixed end vertex. Moreover, if X and τAX are both indecomposable A-modules then the number of arrows Y → X equals the number of arrowsτAX → Y . ♦

132

7.2 The knitting technique

Exercises

7.1.1 Show that if a : L → E and a′ : L → E′ are two source maps, then thereexists an isomorphism ϕ : E → E′ such that a′ = ϕa. Prove the dual statementabout sink maps. This shows that it makes sense to speak of the source map or thesink map.

7.1.2 Show that for some indecomposable A-module M the map 0→M is a sinkmap if and only if M is simple and projective if and only if M = Px for some sinkx of the quiver Q of A, that is, a vertex in which no arrows starts. A vertex inwhich no arrow ends is called a source.

7.1.3 Let y be a vertex of the quiver of A. We decompose radPy into indecom-posable direct summands: radPy =

⊕R∈ΓA

RdR for some integers dR ≥ 0, almostall of them zero. Show that for each R ∈ ΓA the entries of the canonical inclusionRdR → Py yields a basis of the space radA(R,Py)/ rad2

A(R,Py).


In the following we describe the mechanism of the procedure of knitting byconstructing source and sink maps inductively.

For this we define for indecomposable A-modules X,Y the number

δX,Y = dimK

(radA(X,Y )/ rad2

A(X,Y )).

Note that δX,Y is precisely the number of arrows X → Y in the AR-quiverΓA of A. Moreover, for each indecomposable A-module X we denote

X← = Z ∈ ΓA | δZ,X > 0

andX→ = Z ∈ ΓA | δX,Z > 0.

It follows from the existence of source and sink maps given in Proposition 7.2and the characterization given in Lemma 7.1 that these sets are finite.

For Y = Py, the projective indecomposable associated to the vertex y, thenumber δX,Py is the maximal power ` such that X` is a direct summand ofradPy. We define the homomorphism

αX,Py =

α1...

αδX,Py

: X → PδX,Pyy ,

133

7 Knitting

where αi : X → Py is the composition of inclusion in the i-th coordinateX → XδX,Py with the canonical inclusions XδX,Py → radPy → Py.

The next result is the basic tool for knitting. It shows how to obtain a sourcemap starting with a sink map (and vice versa).

Proposition 7.4. Let A be a finite-dimensional K-algebra and X an inde-composable A-module. Let

ρX =[· · · ρZ,X · · ·

]Z

:⊕Z∈X←

ZδZ,X −→ X

be the sink map for X and further for each indecomposable Z ∈ X← let

σZ =

...

σZ,E...

E

: Z −→⊕E∈Z→

EδZ,E

be the source map for Z. For each non-injective summand Z ∈ X← let

µτ−AZ=[· · · µE,τ−AZ · · ·

]E

:⊕E∈Z→

EδZ,E −→ τ−AZ,

be the cokernel of σZ and define

µ>X,τ−AZ

=

µ1...

µδZ,X

: X −→ (τ−AZ)δZ,X ,

where µ1, . . . , µδZ,X are the entries of µX,τ−AZ, that is,

µX,τ−AZ=[µ1 · · · µδZ,X

]: XδZ,X −→ τ−AZ.

Then, the homomorphism

ξX =

...µ>X,τ−AZ

...αX,Py

...

Z,y

: X −→⊕Z∈X←ni

(τ−AZ)δZ,X ⊕

⊕y∈Q0

PδX,Pyy (7.2)

is a source map for X, where X←ni = Z ∈ X← | Z is not injective.

134


Proof. If Y ∈ X→ is indecomposable and non-projective then Z = τAYbelongs to X← and is not injective. Furthermore, by considering the AR-sequences ending in Y we see by Lemma 7.1 and Proposition 7.2 that theentries of µX,Y induce a basis of the space rad(X,Y )/ rad2(X,Y ).

Now, if Y is projective, then ι : radY → Y is a sink map by Proposition 7.2.Thus if we decompose radY into indecomposables radY =

⊕R∈ΓA

RδR,Y ,

we get homomorphisms ιX : XδX,Y → Y whose entries yield a basis for thespace rad(X,Y )/ rad2(X,Y ) by Lemma 7.1.

Thus by Lemma 7.1 the map ξX is a source map for X.

We give a small example to exhibit the use of Proposition 7.4.

Example 7.5. Consider the path algebra A = K−→A 3. The vertex 3 is a

sink of the quiver Q of A thus 0 = radP3 and 0→ P3 is a sink map for P3.Furthermore radP2 = P3. Thus by Proposition 7.4 the source map for P3

is a : P3 → P2, that is, the inclusion of the radical. By Proposition 7.2 thisis also the sink map for P2. Moreover P2 = radP1 and once again by the

same result, we get that the source map for P2 is of the form

[bc

]: P2 →

τ−AP3 ⊕ P1. Therefore we can calculate its cokernel and get a sink map[d e]

: τ−AP3 ⊕ P1 → τ−AP2. Note that we also obtain that b : P2 → τ−AP3

is a sink map for τ−AP3 and therefore e : τ−AP3 → τ−AP2 a source map forτ−AP3. Now P1 and τ−AP2 are both injective, thus the Proposition 7.4 yieldsthat the source map for τ−AP2 is obtained as cokernel of e and has the formf : τ−AP2 → τ−2

A P3.

P3

P2

P1

τ−AP3

τ−AP2

τ−2A P3 .

@

@@@R

@@@@R

@@@@R

a b

c

d

e

f

τA τA

τA

Altogether we have knitted the Auslander-Reiten quiver ΓA as shown in thepicture above. ♦

135

7 Knitting

Exercises

7.2.1 Formulate the dual of Proposition 7.4 and use it to knit ΓA of Example 7.5starting with the injectives.

7.3 Knitting with dimension vectors

Knitting with modules is still rather complicated since cokernels have to becomputed. A reduction of the information yields a far better way to knitin many interesting examples. For this we consider the dimension vectorsof the modules involved in the computations considered as representationsof the quiver of the algebra, see Section 2.2. This has the advantage thatthe computation of cokernels is replaced by the calculation of its dimensionvector as a difference of known vectors using the fact that AR-sequences areshort exact sequences and hence the sum of dimension vectors of the endterms equals the dimension vector of the middle term; see Exercise 5.1.1.We exhibit this in an example.

Example 7.6. We consider the algebra A = KQ/I, where

Q :

@@I

@@I

5 4

3

2

1α

β

γ

δε and I = 〈γα− δβ, εδ〉. (7.3)

We start by calculating the indecomposable projective and the indecompos-able injective A-modules, but we indicate only their dimension vectors:

dimP1 = 011

11, dim I1 = 00

0

01,

dimP2 = 010

10, dim I2 = 00

0

11,

dimP3 = 111

00, dim I3 = 00

1

01,

dimP4 = 110

00, dim I4 = 01

1

11,

dimP5 = 100

00, dim I5 = 11

1

00.

Notice that I5 = P3 and I4 = P1. Since P5 is simple its radical is zero andtherefore there is no non-zero radical homomorphism ending in P5. Further,we have radP4 = P5 and therefore the inclusion P5 → P4 is the sink mapfor P4 and the source map for P5. We have an Auslander-Reiten sequenceP5 → P4 → τ−AP5 and by inspecting the dimension vectors we conclude that

dim τ−AP5 = 110

00 − 10

0

00 = 01

0

00. Consequently τ−AP5 is the simple S4.

136

7.3 Knitting with dimension vectors

Similarly, radP3 = P4 and therefore we have an Auslander-Reiten sequence

P4 → S4 ⊕ P3 → τ−AP4 where dim τ−AP4 = 011

00. So far we have constructed

the part shown in the next picture on the left:

100

00

110

00

111

00

010

00

011

00

@

@R

@@R

= P3

= S4 100

00

110

00

111

00

010

00

011

00

@@R

@@R

@@R

@@R

010

10

011

10

011

11

1

= P2

= P1

The dotted lines join the end terms of the Auslander-Reiten sequences wehave constructed so far and the morphisms are exactly the ones appearingin those two sequences. Proceeding we first observe that P3 = I5 is not thebeginning of an Auslander-Reiten sequence. Further, the simple module S4

is the radical of P2 and hence we know that S4 → P2 is the sink map for

P2. Thus we can determine the dimension vector dim τ−AS4 = 011

10. From

theory we know that τ−AS4 is indecomposable so, there is only one choice:the radical of P1. The result is depicted in the previous picture on the right.

Now, no surprise can occur, since we already have found all indecomposableprojectives, that is, we just have to go on and calculate the Auslander-Reitensequences starting in any non-projective.

100

00

110

00

111

00

010

00

011

00

010

10

011

10

011

11

000

10

001

00

001

11

001

01

000

11

000

01

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

1 PPq

The previous picture shows the Auslander-Reiten quiver of A obtained byknitting. ♦

Example 7.6 shows that knitting with dimension vectors is both: easy andtremendously powerful. It enabled us to produce easily all indecomposablemodules, since for each dimension vector appearing only one possible choice(up to isomorphism) for an indecomposable module is possible.

137

7 Knitting

However, the attentive reader should have noted that it is not clear thatwe found all indecomposables with the knitting procedure. We only knowthat the constructed part is a component of the AR-quiver ΓA since we haveconstructed all arrows starting or ending in any of the constructed modules.But there could be many undetected components out there. As to that thenext result will help us out.

Theorem 7.7 (Auslander). If A is a connected algebra such that its Aus-lander-Reiten quiver ΓA contains a finite component C, then C = ΓA.

Proof. Let M be the direct sum of all (indecomposable) A-modules in C andB = EndA(M) the finite-dimensional K-algebra which by Corollary 3.21 hasa nilpotent radical given by the radical homomorphisms in C. Hence thereis an upper bound, say d > 0, for which gd . . . g1 = 0 for any radicalhomomorphisms g1, . . . , gd in C.

We show that this implies that for any X ∈ ΓA \C and any Y ∈ C we haveHomA(X,Y ) = 0. Suppose that f : X → Y is a non-zero homomorphism.Then f factors through the sink map πY : E → Y ending in Y . Therefore,there is at least one indecomposable direct summand Y1 ∈ C of E and tworadical homomorphisms f1 : X → Y1 and g1 : Y1 → Y such that g1 f1 6= 0.Iterating the argument for f1, we obtain an indecomposable Y2 ∈ C andfurther two radical homomorphisms f2 : X → Y2 and g2 : Y2 → Y1 such thatg1 g2 f2 6= 0. Hence rad2(Y2, Y ) 6= 0. Inductively we find objects Yn ∈ Csuch that radn(Yn, Y ) 6= 0 for any n, in contradiction to radd(Yd, Y ) = 0.Similarly, we have HomA(Y,X) = 0 for any Y ∈ C and any X ∈ ΓA \ C.

Now it follows from the existence of projective covers, Corollary 4.29, that fora fixed Y ∈ C there exists a projective indecomposable module P admittinga non-zero homomorphism P → Y . Hence by the above P lies in C. Since Ais connected, inductively all other projective indecomposable modules belongto C. But then again, for any indecomposable A-module X there exists anon-zero homomorphism Q → X for some projective indecomposable A-module Q and therefore X lies in C.

Exercises

7.3.1 Knit the AR-quiver of the two subspace algebra A = KQ by using dimensionvectors.

7.3.2 Knit the AR-quiver of the three-subspace algebra A = KQ, where Q is thefollowing quiver

138

7.4 Limits of knitting

1 2 3

4

@@R ?

and compare the effort needed now with the one necessary to do Exercise 1.3.3.

7.3.3 Knit the AR-quiver of the algebra A = KQ/I, where the quiver Q and theideal I are given as follows

Q : 1

2

3 4

@@R

-

α

β

γand I = 〈γα〉.

7.3.4 Determine the AR-quiver of B = KQ/J , where Q is the quiver of the Exer-cise 7.3.3 but the ideal is J = 〈γα, γβ〉. You may do this either by knitting or bythe idea explained in Section 3.8, that is, by considering B = A/〈γβ〉.


Knitting is the promised tool which effectively can replace the matrix prob-lem approach – at least in some cases. It is really “self correcting” since anerror in the calculation does not stand long undetected. Unfortunately it isnot that universally applicable as we see in the forthcoming considerations.The orbits under the action of the Auslander-Reiten translate τA are calledAuslander-Reiten orbits or τA-orbits for short. Clearly, in each inde-composable projective such a τA-orbit starts and in each indecomposableinjective a τA-orbit ends. So the question arises whether there might beother orbits at all. For this we work out another example.

Example 7.8. Let A = KQ/I, where

Q :

3

2

1 @@I

αβ

γ

and I = 〈βα〉.

We calculate the indecomposable projectives and their radicals. We obtain:radP3 = 0, radP2 = P3 and radP1 = P3 ⊕ S2. Thus we may start knittingwith the inclusions P3 → P2 and P3 → P1 and obtain the part shown on theleft in the following picture.

139

7 Knitting

100

110

111

010

121

@@R

@@R

= S2

= P1

100

110

111

010

121

101

111

011

001

010

@@R

@@R

@@R

@@R

@@R

@@R

However, it is not clear whether the inclusion ι : S2 → P1 is the source mapfor S2 and we cannot proceed until this question is not settled. So, supposethat f : S2 →M is a radical non-zero homomorphism. Then f2 : K →M2 isnot zero, f2(1) = m 6= 0. Therefore Mβ(m) = Mβ f2(1) = f1 (S2)β(1) = 0since (S2)β = 0. We have m ∈ ImMα because otherwise Km ⊂ M is asubmodule isomorphic to S2 and f a section. So, there exists x ∈ M1 suchthat Mα(x) = m. Now, let g : P1 → M be such that g1(1) = x and notethat by the Yoneda Lemma 4.24 this defines g uniquely. Clearly, we havef = g ι. Thus, we really can proceed knitting and obtain the Auslander-Reiten quiver on the right above.

Observe that two dimension vectors, namely 010 and 1

11 occur twice in

the picture. However, there is only one module with dimension vector 010 ,

namely the simple S2. The two encircled vertices are thus the same andhave to be identified. It follows that τ2

AS2 = S2 and therefore the two A-modules S2 and τAS2 form a τA-orbit not ending in an injective nor beginningin a projective. The case is different with the second dimension vector111 . It clearly occurs as dimension vector of P1, but also as the one of I3.

But I3 and P1 cannot be isomorphic, since (P1)β = 0 and (I3)β 6= 0, sothe second occurrence could be the one coming from I3. If τ−AP1 was notinjective, then τ−2

A P1 would be indecomposable, but if we proceed knittingwe obtain negative numbers which clearly is absurd for a dimension vectorhence τ−AP1 = I3. ♦

Remember, that we reduced to dimension vectors since knitting is so easywith them. Example 7.8 shows the drawback of this reduction: we have toargue in order to understand which modules correspond to the dimensionvectors we found. As we see, knitting sometimes does not work so smoothlyand some ad hoc arguments have to be used. A strategy to attack thisproblem is using coverings, see for example [5]. Unfortunately this is beyondthe scope of this notes. The next example shows a different problem ofknitting.

140


Example 7.9. Let A = KQ, where Q is the quiver (7.3) used in Exam-ple 7.6. Since there are no relations the projectives and their radicals haveto be computed afresh. It turns out that the radical of each projective A-module is projective again. So knitting becomes really easy. We show theresulting part of the Auslander-reiten quiver after some steps in the knittingproces in the next picture, where the top and the bottom row have to beidentified.

P5 =P4 =

P3 =

P2 =

P1 =

P1 =

100

00

110

00

111

00

110

10

221

11

221

11

010

00

121

10

231

21

232

11

111

10

242

21

242

21

453

32

464

32

463

42

342

22

685

53

685

53

7116

64

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

@@R

1 1 1 1PPq PPq PPq PPq

q q q

q q q

It seems that this will go on forever. But how can we be shure about this?It could be that after several thousand steps the dimension vectors startto decrease and finally the injectives appear? If we only look at the totaldimension of the modules only, then the beginning of the AR-quiver looksas shown in the next picture on the left.

12

3

3

7

7

15

9

9

4

11

11

17

@@R @@R @@R

@@R @@R

@@R @@R

@@R @@R

1 1 1PPq PPq PPq

q q q

q q qa

b

c

c

d

d

a′b′

c′

c′

d′

d′

@@R @@R

@@R @@R

@@R

@@R

1 1PPq

Observe that we have an obvious symmetry with respect to the horizontal

141

7 Knitting

axis. We suppose that at some stage ` the dimensions are

a = dimK τ−À P5,

b = dimK τ−À P4,

c = dimK τ−À P3 = dimK τ

−À P2,

d = dimK τ−À P1,

see the previous picture on the right for illustration. We denote the dimen-sions for the τ−A -translated modules by a′, b′, c′ and d′ respectively. Usingthat the Auslander-Reiten sequences are short exact sequences we get thefollowing relations:

a′ = b− a,b′ = 2c+ a′ − b = 2c− a,c′ = d+ b′ − c = d+ c− a,d′ = 2c′ − d = d+ 2c− 2a.

We show that for all ` the following inequalities hold:

d > c > b ≥ 2a. (7.4)

Clearly they are valid for ` = 0. We suppose they hold for some ` andcalculate

d′ − c′ = d+ 2c− 2a− (d+ c− a) = c− a > 0

hence d′ > c′. Similarly the inequalities c′ > b′ ≥ 2a′ can be shown. Thusby induction (7.4) hold for all `. Furthermore the argument shows also thata′ ≥ a, b′ > b, c′ > c and d′ > d. Consequently the dimensions proceed toincrease forever. A similar picture is obtained by knitting in the oppositedirection starting with the injective indecomposables to the left.

Are this all modules? Clearly not, since the simples S3 and S2 do notappear. Let us look at S3. Notice that f : P4 → P3 is a minimal projectivepresentation for S3, hence τAS3 is the kernel of νAf : I4 → I3, where νA isthe Nakayama functor. Therefore Ker(νAf) has dimension vector 01

0

11. We

conclude that P5 ⊕ P3 → P1 is the minimal projective presentation of τAS3

and we calculate dim τ2AS3 = 11

2

12. Similarly, we can calculate τ−AS3, τ

−2A S3

and thus the sequence

112

12 01

0

11 00

1

00 11

0

11 01

0

10

142

7.5 Preprojective components

is obtained. ♦

Example 7.9 shows two problems: First, knitting may continue forever or itmay stop, but the process itself does not tell us which one occurs in a givenexample. Second, as we have seen in the calculation of the τA-translates ofS3 that we really have to know the module structure for calculating the pro-jective presentations, the dimension vector is not enough. Thus we cannotknit as we would like to.

Exercises

7.4.1 Determine the AR-quiver of A = KQ/I, where the quiver Q and the ideal Iare as follows.

Q : 1 2

34

-

?

6

α

γ

βδ and I = 〈βα, γβ, δγ, αδ〉.

7.4.2 Let A = KQ, where Q is the four subspace quiver:

1 2 3 4

5

@@RBBBN

Start knitting from the projectives. Prove that the component you are knittingis infinite. Do the same with the component obtained by knitting from the injec-tives and prove that there is some indecomposable which does not belong to eithercomponent.


A component C of the Auslander-Reiten quiver ΓA of a finite-dimensionalalgebra A is called preprojective or sometimes postprojective if each τA-orbit contains a projective indecomposable (and consequently there are onlyfinitely many τA-orbits) and there is no cycle in C. Dually, a componentwithout cycle in which each τA-orbit contains an injective indecomposableis called preinjective (nobody calls them “postinjective”). A componentwithout any projective or injective A-module is called regular. An A-module contained in a preprojective, resp. preinjective, resp. regular com-ponent is called preprojective, resp. preinjective, resp. regular.

143

7 Knitting

Knitting of a preprojective component C is always successful: if we startwith a simple projective A-module in it, either the knitting process goes onforever for some τA-orbits or in each τA-orbit an injective indecomposableA-module is reached and in that case the component C is finite and henceC = ΓA by Theorem 7.7.

Note that not every algebra admits a preprojective component. For instancethe algebra considered in Example 7.8 has only one component which con-tains a cycle and therefore is not preprojective. However, there are niceclasses of algebras which do have a preprojective and a preinjective com-ponent. For this, it will be useful to systematize many examples which wealready have considered.

An algebra A is called hereditary if each submodule of a projective A-module is projective again. Note that this happens if and only if eachA-module has a projective presentation 0→ Q1 → Q0, that is, if and only ifExt2

A(M,N) = 0 for any two A-modules M and N . Now, this last conditioncan also be stated in different terms: it means that each A-module has aninjective presentation J0 → J1 → 0, in other words, that each quotient ofan injective A-module is injective again. A different description is given inExercise 7.5.1: a basic and finite-dimensional K-algebra A is hereditary ifand only if it is isomorphic to the path algebra KQ of a quiver Q withoutcycle.

Proposition 7.10. If A is a hereditary and finite-dimensional K-algebra,then in ΓA each projective A-module is contained in a preprojective compo-nent and dually each injective A-module in a preinjective component. If A isconnected, then there exists a unique preprojective component and a uniquepreinjective component. Moreover in that case, the preprojective componentis preinjective if and only if A is of finite representation type.

Proof. Note that without loss of generality, we may assume A to be basic, seeProposition 3.16. Hence by Exercise 7.5.1, we have A = KQ, for some quiverQ. Since A is finite-dimensional the quiver Q has no cycle, see Exercise 2.4.4.

LetQ(0) be the full subquiver of ΓA obtained by the projectives, that is, thequiver with vertices the projective indecomposableA-modules in ΓA togetherwith all the arrows in ΓA between them. More generally, a subquiver Q′

of a quiver Q is a quiver given by two subsets, that is Q′0 ⊆ Q0 and Q′1 ⊆ Q1

such that sQ′(α) = sQ(α) and tQ′(α) = tQ(α) for each α ∈ Q1.

It follows from the Yoneda Lemma 4.24 that each homomorphism f : Px →Py is given by f = Pa for some a ∈ exAey, where Pa : Px → Py, b 7→ ba.

144


Hence it follows that the arrows in ΓA between projectives can be chosen asPα for an arrow α of the quiver of A. This shows that Q(0) is isomorphic toQop and hence contains no cycle.

Therefore there exists Px ∈ Q(0) which is a source, or equivalently x ∈ Q isa sink. Note that

σx =[· · · Pα · · ·

]α

: Px →⊕

α : y→xPy

is a source map for Px by Proposition 7.4. Hence we obtain τ−APx as cokernelof σx:

ρx =

...

ρx,α...

:⊕

α : y→xPy → τ−APx.

We may assume that the irreducible homomorphisms ρx,α are chosen asarrows in ΓA and that these are all arrows ending in τ−APx.

Denote by Q(1) the quiver obtained by deleting Px from Q(0) and addingτ−APx together with all arrows ρx,α. Note, that Q(1) is isomorphic to thequiver obtained from Q(0) by inverting the direction of all arrows ending inPx. We may now repeat the arguments with every source in Q(0) to obtainQ(t). Let Py ∈ Q(t) be a source. It follows from Proposition 7.4 that thesource map for Py looks like

σy =[· · · Pβ · · · ρx,α · · ·

]β,α

: Py →⊕

β : z→yPz ⊕

⊕α : y→x

τ−APx.

Note that we may proceed that way until no vertex of the original quiverQ(0) is left. Finally, we get a quiver Q(n) which is isomorphic to Q(0) andconsists of the vertices τ−APx for all x ∈ Q0 together with the arrows in ΓAbetween them. Proceeding this way, we either continue forever or at somestep reach an injective A-module. In the latter case all τA-orbits are finite.Thus every connected component of Q yields a component C of ΓA.

Clearly in each such component C there are only paths from τ `Px to τmPy if` ≥ m. This shows that there is no cycle in C and that C is a preprojectivecomponent. The statement about the components containing the injectivesis proved dually.

If A is connected, then the projective indecomposable A-modules are con-nected by irreducible homomorphisms and therefore all projectives in ΓAmust belong to the same component. The statement about the preinjectivecomponent is proved dually.

145

7 Knitting

Proposition 7.11. If C is a preprojective or preinjective component of ΓA,then for each indecomposable A-module X ∈ C we have EndA(X) ' K andExt1

A(X,X) = 0.

Proof. Note that by definition, the arrows in C induce a partial order “≤”in the vertices of C. Thus for X,Y ∈ C we have X ≤ Y if and only ifthere exists a path from X to Y or X = Y . We show that HomA(X,Y ) 6= 0implies X ≤ Y . Indeed, in case C is preprojective, we consider the subquiverC ′ of C obtained by deleting all Z ∈ C for which Z 6≤ Y . Note that C ′ is afinite quiver: first let Y = τ−À Px for some x ∈ Q0 and some ` ≥ 0. Then theelements of the τA-orbit of Y which belong to C ′ are τ−iA Px for i = 0, . . . , `.If Z ∈ Y← then similarly only finitely many elements of the τA-orbit of Zbelong to C ′. Proceeding this way can consider all τA-orbits, each havingonly a finite number of elements in C ′.

We have to show that X ∈ C ′. Now, if f : X → Y is a non-zero homomor-phism, we get that f factors through Y← ⊆ C ′, say f =

∑ti=1 figi where

gi : Zi → Y for some Zi ∈ Y← and fi : X → Zi. Note that Zi ∈ C ′ for alli. If we suppose that X 6∈ C ′, then we can continue factoring f througharbitrarily large number of irreducible homomorphisms, but since C ′ is finitethere is a bound M for which radM (Z, Y ) = 0 for all Z ∈ C ′, a contradiction.This shows that X ∈ C ′ and therefore X ≤ Y .

As a consequence we have HomA(X,E) = 0 if E is the middle term ofthe Auslander-Reiten sequence ending in X and therefore radA(X,X) = 0,which in turn implies EndA(X) ' K. By Theorem 6.16, we have thatdimK Ext1

A(X,X) = dimK HomA(X, τAX) ≤ dimK HomA(X, τAX) = 0since there is no path in C from X to τAX.

Example 7.12. The preprojective, resp. preinjective, resp. regular modulesover the Kronecker algebra are of the form Qn, resp. Jn, resp. Rn,λ, seeSection 6.2. ♦

Exercises

7.5.1 Let A = KQ/I, where Q is an arbitrary quiver and I is an admissible ideal.Let ρ1, . . . , ρt be relations such that I = 〈ρ1, . . . , ρt〉 and assume that t is minimal.Then we have that 〈ρ1, . . . , ρi−1, ρi+1, . . . , ρt〉 6= I for each i = 1, . . . , t. Show thatradPs(ρi) is not projective. This implies that A is not hereditary. Prove converselythat KQ is hereditary for any quiver Q without cycle.

7.5.2 Knit the preprojective component of ΓA for A = KQ, where Q is the follow-ing quiver.

146


Q :

2

1

3 4 5 6

@R- - -

147

7 Knitting

148

Chapter 8

Combinatorial invariants

The reduction from modules to dimension vectors can not only be seen as aloss of information but also as a gain in simplicity. Several useful tools arepresented for working with dimension vectors. The first of them is based onhomological algebra, the second linearizes the Auslander-Reiten translationas far it is possible and the last plays a crucial role for characterizing theindecomposable modules in the reduction process.

8.1 Grothendieck group

We have seen in Section 7.3 that dimension vectors are very useful. We nowgive a more theoretic approach to them. For this we recall some notionsfrom ring theory.

Let R be a ring. An R-module F is said to be free if there exists a subsetS ⊂ F such that for each R-module M and each map ϕ : S → M thereexists a unique homomorphism of R-modules ϕ : F →M such that ϕ|S = ϕ.In that case S is called a basis of F . If the basis S is finite, then F is calledfinitely generated free. Note that this universal property defines thefree R-module uniquely, up to isomorphism, see Exercise 8.1.1. If S is aset then R(S) denotes the R-module of functions f : S → R which are zeroalmost everywhere, that is, s | f(s) 6= 0 is a finite set. Commonly oneidentifies s ∈ S with δs ∈ R(S) where

δs : S → R, t 7→

s, if s = t,

0, else.

149

8 Combinatorial invariants

Then R(S) is a free R-module with basis S, see Exercise 8.1.2. This showsthat for each set S there exists a free R-module with basis S. This explicitconstruction shows also that in a free R-module F with basis S each elementx ∈ S can be written as (finite) linear combination of elements in S withcoefficients in R. A particular interesting case is when R = Z the ring ofintegers. We then speak of free abelian groups.

Temporarily we denote the isomorphism class of an A-module X by cl(X).We know from Lemma 4.14 that the isomorphism classes of A-modules forma set

S = cl(X) | X ∈ modA.

The Grothendieck group K(A) of an algebra A is the quotient

K(A) = Z(S)/I

where I is the ideal generated by all linear combinations

δcl(X′) − δcl(X) + δcl(X′′)

whenever there exists a short exact sequence Y ′ → Y → Y ′′ of A-modulesfor some Y ′ ∈ cl(X ′), Y ∈ cl(X) and Y ′′ ∈ cl(X ′′). The coset of δcl(X) inK(A) is called the class of an A-module X in K(A) and will be denotedby [X].

Since 0 → 0 → 0 is a short exact sequence we have δcl(0) = δcl(0) − δcl(0) +δcl(0) ∈ I and therefore [0] is the neutral element of K(A), denoted by 0.Furthermore, [X] = [Y ] holds if X and Y are isomorphic A-modules since

any isomorphism f : X → Y yields a short exact sequence Xf−→ Y → 0 and

therefore [X] + [0] − [Y ] = 0 n K(A). Moreover [X ′ ⊕ X ′′] = [X ′] + [X ′′]since the split sequence X ′ → X ′ ⊕X ′′ → X ′′ is a short exact sequence.

Remark 8.1. The Grothendiek group is defined as a quotient F/I of a freeabelian group F modulo an ideal I generated by some set of relations R.We denote by π : F → F/I the canonical projection.

Now, every homomorphism ϕ : F/I → Z yields a group homomorphismϕ = ϕπ : F → Z which satisfies ϕ(r) = 0 for each r ∈ R. Conversely, eachhomomorphism ψ : F → Z which satisfies ψ(r) = 0 for all r ∈ R induces agroup homomorphism ϕ : F/I → Z such that ϕπ = ψ.

Therefore, to give a Z-linear map f : K(A)→ Z it is necessary and sufficientto define the values f(M) for each M ∈ modA such that f(M) = f(M ′) +f(M ′′) whenever there exists a short exact sequence M ′ →M →M ′′. ♦

150

8.1 Grothendieck group

Proposition 8.2. If A is a finite-dimensional K-algebra with quiver Q,then

K(A)→ ZQ0 , [M ] 7→ dimM.

is a Z-linear bijection, that is, an isomorphism of abelian groups.

Proof. Let S be the set of isomorphism classes in modA. Define

S → ZQ0 , cl(M) 7→ dimM.

Now, if M ′ →M →M ′′ is a short exact sequence, then dimM = dimM ′ +dimM ′′, see Exercise 5.1.1. Thus, by Remark 8.1 the above map induces alinear function

K(A)→ ZQ0 , [M ] 7→ dimM.

If M is not a simple module, then there exists a submodule M ′ and hence[M ] = [M ′] + [M/M ′]. This shows that for any A-module M the class [M ]is a linear combination of the classes of the simple A-modules [S1], . . . , [Sn],where the coefficient of [Si] is the number of times Si appears as factor in aJordan-Holder filtration of M .

Hence [M ] =∑

i∈Q0ni[Si], where ni is the dimension of M in the vertex i,

viewed as representation of the quiver Q of A. This shows that K(A) isthe free abelian group with basis [Si], i ∈ Q0, that is K(A)→ ZQ0 , [M ] 7→dimM is an isomorphism of abelian groups.

In the forthcoming we will not distinguish between [M ] and dimM . Fur-thermore we denote the class [Si] by ei, that is, the canonical basis vectorof ZQ0 which should not be confounded with the trivial path in i which isdenoted by the same symbol.

Exercises

8.1.1 Let F (resp. F ′) be a free R-module with basis S (resp. S′). Show that if

there exists a bijection ψ : S → S′, then there exists an isomorphism ψ : F → F ′

such that ψ|S = ψ. Hint: use that idF is the unique homomorphism which satisfiesidF j = j, where j : S → F denotes the inclusion.

8.1.2 Show for a ring R and a set S that R(S) is a free R-module with basis S.

8.1.3 Let

0a0−→M1

a1−→M2a2−→ . . .

at−1−−−→Mtat−→ 0

151


be an exact sequence. Show that the alternating sum of the classes in theGrothendieck group vanishes:

t∑i=1

(−1)i[Mi] = 0.

Hint: consider the short exact sequences Ker ai →Mi → Im ai = Ker ai+1.

8.2 The homological bilinear form

Suppose that the finite-dimensional K-algebra A is of finite global dimen-sion. Then for any two A-modules X and Y we define

〈X,Y 〉 =∞∑i=0

(−1)i dimK ExtiA(X,Y ). (8.1)

Note that the sum on the right hand side is finite since gldimA < ∞ andthat Ext0

A(X,Y ) = HomA(X,Y ) by Exercise 5.3.3. The next result showsthat the value of 〈X,Y 〉 depends only on the classes [X] and [Y ] in K(A).

Lemma 8.3. Let A be a finite-dimensional K-algebra of finite global dimen-sion. If X,X ′, Y and Y ′ are A-modules such that [X] = [X ′] and [Y ] = [Y ′],then 〈X,Y 〉 = 〈X ′, Y ′〉 holds. Moreover, the map

〈−,−〉 : K(A)×K(A) −→ Z

induced by (8.1) is bilinear.

Proof. By Theorem 5.20 for each short exact sequence M ′ → M → M ′′ ofA-modules and each A-module L we have a long exact sequence

· · · → Exti(L,M ′)→ Exti(L,M)→ Exti(L,M ′′)→ Exti+1(L,M ′)→ · · · .

Note that in an exact sequence the alternating sum of the dimensions iszero, see Exercise 8.2.1. This shows that

〈L,M ′〉 − 〈L,M〉+ 〈L,M ′′〉 = 0.

By Remark 8.1 the map 〈L,−〉 induces a Z-linear map K(A) → Z. Simi-larly, for every fixed N the map 〈−, N〉 induces a Z-linear map K(A)→ Z.Thus the result follows.

152

8.2 The homological bilinear form

The bilinear form K(A)×K(A)→ Z defined in Lemma 8.3

〈[X], [Y ]〉 =∞∑i=0

(−1)i dimK ExtiA(X,Y ). (8.2)

is called homological bilinear form of A.

Example 8.4. Let A be a hereditary algebra, that is, A = KQ forsome quiver Q by Exercise 7.5.1. For any projective A-module P andany A-module M we have 〈[P ], [M ]〉 = dimK HomA(P,M) since all ex-tension groups are zero. Thus if P is indecomposable, say P = Pi, then〈[Pi], [M ]〉 = dimKMi. Since there is no cycle in Q we therefore have〈[Pi], [Pi]〉 = 1.

For simple A-modules Si and Sj we have

〈[Si], [Sj ]〉 = dimK HomA(Si, Sj)− dimK Ext1A(Si, Sj).

Note that by Exercise 5.3.2 we know that dimK Ext1A(Si, Sj) is the number

of arrows j → i in the quiver Q and by Schur’s Lemma 4.33 we have thatHomA(Si, Sj) = 0 for i 6= j. ♦

Let A be a finite-dimensional K-algebra of finite global dimension and quiverQ with verticesQ0 = 1, . . . , n. We denote by Pi projective indecomposableA-module associated to the vertex i and by pi = [Pi] its class in K(A). Thenthe Cartan matrix CA ∈ Kn×n of A is defined by

(CA)ij = dimK HomA(Pi, Pj) = 〈pi, pj〉.

Lemma 8.5. The Cartan matrix satisfies the following identities

CA[Sj ] = [Pj ],

C>A [Si] = [Ii].

Furthermore, CA is invertible if and only if the classes p1, . . . pn of the in-decomposable projectives are linearly independent in K(A).

Proof. Note that 〈pi, pj〉 = dimK HomA(Pi, Pj) = (pj)i the i-th entry of thevector pj = [Pj ]. Thus the j-th column of the matrix CA is precisely pj .Since HomA(Pi, Pj) ' HomA(Ii, Ij) = [Ii]j the i-th row of CA is [Ii]

>. Thisshows the first two equations.

If the vectors p1, . . . , pn are linearly independent, then CA is certainly invert-ible. To see the converse we assume that CA is invertible and denote by r

153


the rank of the subgroup of K(A) generated by p1, . . . , pn. Then ei = C−1A pi

implies r ≥ n since C−1A p1, . . . , C

−1A pn generate a subgroup of rank at most

r.

Lemma 8.6. Let A be a finite-dimensional K-algebra. If A is of finite globaldimension, then the classes of the indecomposable projective A-modules gen-erate K(A) and are linearly independent.

Proof. If A is of finite global dimension, then each A-module admits a pro-jective resolution which has finite length. It follows thus from Exercise 8.1.3that the class [M ] of an A-module M can be expressed as linear combina-tion of the vectors p1, . . . , pn with integer coefficients. Therefore the vectorsp1, . . . , pn generate K(A) and since K(A) is a free abelian group isomorphicto Zn these vectors must also be linearly independent.

Remarks 8.7 (a) The statement of Lemma 8.6 can be reformulated asfollows: under the given hypothesis the Cartan matrix CA of A is invertible.

(b) The converse of Lemma 8.6 is wrong: it is easy to construct algebrasof infinite global dimension but with invertible Cartan matrix, see Exer-cise 8.2.2.

(c) A note of caution: K(A) is a free abelian group of rank n, that is,it is isomorphic to Zn as abelian group. If x1, . . . , xn generate Zn, thenthese vectors are linearly independent, but the converse is false. Considerthe case where xi = 2ei for all i. Then the vectors x1, . . . , xn are linearlyindependent, but they do not generate Zn since there exist vectors, forexample ei, which cannot be expressed as linear combination of x1, . . . , xnwith integer coefficients. We can reformulate these conditions using thematrix M whose i-th column is xi. To say that x1, . . . , xn are linearlyindependent is equivalent to detM 6= 0, whereas x1, . . . , xn generate Zn ifand only if detM = ±1. ♦

Proposition 8.8. Let A be a finite-dimensional K-algebra. If A is of finiteglobal dimension, then

〈x, y〉 = x>C−>A y

holds for all x, y ∈ K(A), where C−>A = (C−1A )>.

154

8.3 The Coxeter transformation

Proof. Since gldimA < ∞ it follows by Lemma 8.6 that CA is invertible.Consequently ei = C−1

A [Pi] and we see

〈pi, pj〉 = e>i pj = p>i C−>A pj

holds for all i, j. Since p1, . . . , pn generate K(A) the equation 〈x, y〉 =x>C−>A y follows for all x, y ∈ K(A).

Exercises

8.2.1 Show the following generalization of Exercise 5.1.1: if 0 → M1 → M2 →. . .→ Mt → 0 is an exact sequence of A-modules, then the alternating sum of thedimension vanishes:

t∑i=1

(−1)i dimKMi = 0.

8.2.2 Let A = KQ/I, where the quiver Q is as shown in the following picture andthe ideal I is generated by the relation βαβα.

Q : 1 2j

Y

α

β

Calculate the Cartan matrix CA and verify that CA is invertible. Calculate theprojective resolution of the simple S1 to see that gldimA =∞.


We suppose that the finite-dimensional K-algebra A is of finite global dimen-sion. For each A-module M we get two Z-linear maps K(A) → Z, namely〈[M ],−〉 and 〈−, [M ]〉. In this section we are looking for a Z-linear mapK(A) → K(A) which sends [X] to [τAX]. Note that one cannot expect[X] 7→ [τAX] to be a well-defined map on K(A), see Exercise 8.3.1.

We define the Coxeter transformation as the unique linear map K(A)→K(A) which satisfies ΦA[Pi] = −[Ii]. The matrix representing ΦA in thecanonical basis is called Coxeter matrix and denoted also by ΦA.

Lemma 8.9. For a projective presentation f : Q→ P of an indecomposableA-module M the following equation holds

[τAM ] = ΦA[M ]− ΦA[Ker f ] + [νAM ].

155


Proof. By definition we have [νAQ] = −ΦA[Q] and [νAP ] = −ΦA[P ]. Weapply Exercise 8.1.3 to the exact sequence

0→ Ker f → Qf−→ P →M → 0

and get [P ] − [Q] = [M ] − [Ker f ]. Applying Exercise 8.1.3 to the exactsequence (6.23):

0→ τAM → νAQνAf−−→ νAP → νAM → 0.

we can calculate the class of τAM

[τAM ] = [νAQ]− [νAP ] + [νAM ]

= ΦA[P ]− ΦA[Q] + [νAM ]

= ΦA[M ]− ΦA[Ker f ] + [νAM ]

which is what we wanted to prove.

That the formula of Lemma 8.9, valid for all A-modules M , can take aparticularly nice form in special cases is shown in the next result.

Proposition 8.10. Let A be a finite-dimensional K-algebra of finite globaldimension and M an indecomposable A-module. If HomA(M,A) = 0 andpdimM ≤ 1, then ΦA[M ] = [τAM ].

Proof. Since pdimM ≤ 1 there exists a projective presentation f : Q → Pwhich is injective. Thus Ker f = 0 and HomA(M,A) = 0 implies thatνAM = D HomA(M,A) = 0. Hence [τAM ] = ΦA[M ] by Lemma 8.9.

The next result shows how the dimension vectors of preprojective or prein-jective modules over finite-dimensional hereditary K-algebras can be calcu-lated.

Corollary 8.11. If A = KQ is a finite-dimensional hereditary K-algebra,then [τ−À Pi] = Φ−`[Pi] holds for each i ∈ Q0 and each ` > 0 such thatτ−`Pi 6= 0 and similarly [τ ÀIi] = Φ`[Ii] holds for each i ∈ Q0 and each ` > 0such that τ Ìi 6= 0.

Proof. Let M be an indecomposable A-module. If HomA(M,A) 6= 0, thenM is projective since A is hereditary. Therefore, if M = τ−À Pi for ` > 0,then HomA(M,A) = 0. Hence by Proposition 8.10 we have [τAM ] = ΦA[M ].Inductively we get [Pi] = [τ ÀM ] = Φ`

A[M ] thus Φ−À [Pi] = [M ] = [τ−À Pi].Similarly, if M = τ ÀIi is not projective, then [τAM ] = ΦA[M ] again byProposition 8.10. Therefore [τ ÀIi] = Φ`

A[Ii] follows inductively.

156


Lemma 8.12. If the finite-dimensional K-algebra A has finite global di-mension, then ΦA = −C>AC

−1A and

〈x, y〉 = −〈y,ΦAx〉 = 〈ΦAx,ΦAy〉 (8.3)

holds for all x, y ∈ K(A).

Proof. Since pi = CAei and [Ii] = C>Aei we have

ΦApi = −[Ii] = −C>Aei = −C>AC−1A pi

and therefore ΦA = −C>AC−1A .

Now, by Proposition 8.10

−〈y,Φx〉 = y>C−>A C>AC−1A x = y>C−1

A x = x>C−>A y = 〈x, y〉

and thus the first equation of (8.3) follows. The second equation of (8.3)follows by applying twice the first.

Example 8.13. Let A be the Kronecker algebra. The Auslander-Reitenquiver was determined in Section 6.2 but we were not able to determine theAR-translate τAM for those indecomposable modules M which lie neitherin the preinjective nor in the preprojective component, that is M = Rn,λfor some n > 0 and λ ∈ K ∪ ∞, see (6.1). The dimension vector of M is[M ] = (n, n) ∈ Z2 which is an eigenvector of ΦA, see Exercise 8.3.2. SinceHomA(Rn,λ, A) = 0, see Section 6.2, we may apply Proposition 8.10 andget [τARn,λ] = ΦA[Rn,λ] = [Rn,λ]. Because τARn,λ is indecomposable wemust have τARn,λ = Rn,µ for some µ ∈ K ∪ ∞. The Auslander-Reitensequence ending in Rn,λ yields non-zero morphisms Rm,µ → E → Rn,λ andtherefore µ = λ follows from the description of the morphism spaces givenin Section 6.2. This shows that

τARn,λ = Rn,λ

for all n > 0 and λ ∈ K ∪∞ completing the description of the Auslander-Reiten translate for the Kronecker algebra. ♦

Exercises

8.3.1 Show that [X] 7→ [τAX] is not a well-defined map. For this let A = K−→A 2

and let X = P1, Y = S1 ⊕ S2. Show that [X] = [Y ] but [τAX] 6= [τAY ].

8.3.2 Calculate the Coxeter transformation ΦA = −C>AC−1A for the Kronecker

algebra A. Show that 1 is the unique eigenvalue of ΦA and that the eigenvectors xwith respect to this eigenvalue satisfy x1 = x2.

157


8.4 Quadratic forms

Let A be a finite-dimensional K-algebra. If A is of finite global dimension,then the quadratic form associated to the homological bilinear form

χA : K(A)→ Z, x 7→ 〈x, x〉 (8.4)

is called Euler form or sometimes homological form.

Note that χA can only be defined if A is of finite global dimension, sinceotherwise the right hand side of (8.2) is not a finite sum. If A is not of finiteglobal dimension, we still can define a bilinear form 〈−,−〉T and from thisthen a quadratic form by truncating the Euler from in the following way:set

〈[Si], [Sj ]〉T = dimK HomA(Si, Sj)− dimK Ext1A(Si, Sj)

+ dimK Ext2A(Si, Sj)

(8.5)

for simple A-modules Si, Sj and extend it linearly to obtain a bilinear formon K(A). The associated quadratic form qA : K(A) → Z, x 7→ 〈x, x〉T isgiven by

qA(x) =∑i∈Q0

x2i −

∑(i→j)∈Q1

xixj +∑i,j∈Q0

xixj dimK Ext2A(Si, Sj). (8.6)

It is called Tits form, sometimes geometric form or Ringel form.

Remark 8.14. If gldimA = ∞ then χA is undefined. If gldimA ≤ 2 wehave χA = qA since we have

〈ei, ej〉 =∞∑i=0

dimK ExtiA(Si, Sj) =2∑i=0

dimK ExtiA(Si, Sj) = 〈ei, ej〉T

for all i, j ∈ Q0. If the global dimension of A is finite but larger than 2 thenχA and qA are both defined but do not have to be equal. ♦

Note that both quadratic forms χA and qA have the following shape:

q : Zn → Z, q(x) =∑n

i=1qix

2i +

∑i<j

qijxixj (8.7)

for some integer coefficients qi, qij . If qi = 1 for all i then q is called unitform.

158

8.4 Quadratic forms

If q : Zn → Z is a unit form, we define its graph ∆(q) to have 1, . . . , n asvertices and edges as follows: between the vertices i and j there are |qij | fulledges if qij < 0 and |qij | dotted edges if qij > 0. Sometimes multiple edgesare indicated by labels written close to a single edge. Conversely, if G is agraph with full and dotted edges but without loops nor mixed edges, thatis, a pair of vertices (i, j) such that there is a full and a dotted edge, thenwe associate to G the unit form p = q(G) such that G = ∆(p).

If Q is a quiver then the underlying graph of Q is the graph having thesame vertices as Q and aij +aji full edges joining the vertices i and j, whereaji is the number of arrows from i to j. It is customary to say that theunderlying graph is obtained “by forgetting the orientation” of the arrowsin Q.

We start by investigating very special cases of unit forms.

Example 8.15. Let q be a unit form such that its graph ∆(q) is a Dynkindiagram, that is, one of the graphs of the following list:

r r r r1 2 n− 1 n

An (n ≥ 1): p p p

r r r r rr

1 3 4 n− 1 n

2

Dn (n ≥ 4): p p p

En (n = 6, 7, 8): r r r r r rp p pr

1 2 4 5 n− 1 n

3

(8.8)

Note that in these examples there is no dotted edge. Moreover, there are twoinfinite families and three cases E6, E7 and E8 which are called exceptional.These graphs play a fundamental role in many classification problems inmodern algebra. Nowadays it is common to speak of Dynkin diagrams eventhough these structures first appeared at the end of the 19th century whenWilhelm Killing and Elie Cartan classified semisimple Lie algebras over thecomplex numbers. ♦

We observe that the linearly oriented quiver−→An has An as underlying graph.

Now we compare two unit forms p : Zm → Z and q : Zn → Z. For this weassume T ∈ Zm×n to be some integer matrix such that p = q T . If n = mand T is Z-invertible, T ∈ GLn(Z), then we say that p and q are equivalent

159


and denote it by p ∼ q. In the particular case where T is a permutationmatrix, that is, Tei = eσ(i) for some permutation σ ∈ Sn, then we say thatp and q are isomorphic.

There are also interesting cases for p = q T for T ∈ Zn×m not a squarematrix: If m ≤ n and each column of T contains precisely one entry 1 andeach row at most one entry 1 whereas all other entries are zero, then pis called a restriction of q. Note that this happens precisely when ∆(p)is obtained from ∆(q) by deleting some of its vertices (and relabeling theremaining ones).

We recall from linear algebra that by Sylvester’s law of inertia a quadraticform q : Rn → R is, up to equivalence, uniquely defined by the indicesof inertia: q is equivalent to a form p : Rn → R which is given by p(x) =∑n

i=1 pix2i . The number n+ (resp. n−) of positive (resp. negative) coefficients

pi is called positive index of inertia, resp. negative index of inertia andboth indices are uniquely determined by q. The difference n − n+ − n− iscalled corank of q.

We can apply these concepts to a unit form q : Zn → Z by defining qR : Rn →R by qR(x) = q(x) for all x ∈ Zn. That qR is uniquely defined by thisfollows from the fact that the coefficients qi and qij in the expression (8.7)are uniquely defined by qi = q(ei) and qij = q(ei + ej) − q(ei) − q(ej).Since a unit form satisfies q(ei) = 1 we conclude n+ > 0. Hence there arethree important cases to be considered: if n+ = n, then q is said to bepositive definite. If n− = 0 but possibly n > n+ then q is called positivesemidefinite. The third case is when n− > 0 and in that case q is calledindefinite.

Lemma 8.16. Let q : Zn → Z be any quadratic form. Then the followingcharacterizations hold.

(a) The form q is positive definite if and only if q(x) > 0 for all x ∈ Zn\0if and only if qR(x) > 0 for all x ∈ Rn \ 0.

(b) The form q is positive semidefinite if and only if q(x) ≥ 0 for allx ∈ Zn if and only if qR(x) ≥ 0 for all x ∈ Rn.

Proof. First, we observe that q is positive semidefinite if and only if qR(x) ≥0 for all x ∈ Rn and that this implies that q(x) ≥ 0 holds for all x ∈ Zn.Similarly q is positive definite if and only if qR(x) > 0 for all x ∈ Rn \ 0and this implies that q(x) > 0 for all x ∈ Zn \ 0.

160

8.4 Quadratic forms

Assume now that q(x) ≥ 0 for all x ∈ Zn and denote by qQ : Qn → Q theobvious extension of q. Since qQ(ab ) = 1

b2q(a), we see that qQ(x) ≥ 0 for all

x ∈ Qn and by continuity we have qR(x) ≥ 0 for all x ∈ Rn. This proves(b). To see (a) it remains to show that if there exists some non-zero x ∈ Rnwith qR(x) = 0, then there exists some non-zero z ∈ Zn with q(z) = 0.

Indeed, if qR(x) = 0 for some non-zero x, then qR attains in x a globalminimum. Therefore

∂qR∂xi

(x) = 0 (8.9)

for all i = 1, . . . , n. But that means that x satisfies the system of linearequations (8.9) non-trivially. Since all coefficients in (8.9) are rational thereexists also a non-trivial solution y in Qn. That is qQ(y) = 0. By multiplyingy with the least common multiple m of the denominators we obtain a non-zero vector z = my ∈ Zn with q(z) = 0.

Lemma 8.17. Each unit form q whose graph ∆(q) is a Dynkin diagram ispositive definite.

Proof. We start by doing this for the case ∆(q) = Dn. Then we have

q(x) =x21 − x1x3 + x2

2 − x2x3 + x23 − x3x4 + x2

4 − . . .− xi−1xi + x2

i − . . .− xn−1xn + x2n

=(x1 − 12x3)2 + (x2 − 1

2x3)2 + 12(x3 − x4)2 + . . .

+ 12(xi − xi+1)2 + . . .+ 1

2(xn−1 − xn)2 + 12x

2n.

This shows that q(x) > 0 whenever x 6= 0 and hence q is positive definite if∆(q) = Dn.

Clearly, the diagram An can be obtained as restriction of Dn+1 and a restric-tion of a positive definite quadratic form is again positive definite. Thereforeq is positive definite if ∆(q) = An.

Finally, we study the case ∆(q) = E8. Then we have

q(x) =x21 − x1x2 + x2

2 − x2x4 + x23 − x3x4 + x2

4 − . . .− xi−1xi + x2

i − . . .+ x28

=(x1 − 12x2)2 + 3

4(x2 − 23x4)2 + (x3 − 1

2x4)2 + 512(x4 − 6

5x5)2

+ 25(x5 − 5

4x6)2 + 38(x6 − 4

3x7)2 + 13(x7 − 3

2x8)2 + 14x

28

which shows that q is positive definite if ∆(q) = E8. Since E7 and E6 areboth restrictions of E8, we get the desired result.

161


We also give examples of unit forms which are not positive.

Example 8.18. The graphs of the following list are called extendedDynkin diagrams, by some authors they are called Euclidean diagrams.

A1:

r?1

1

An (n ≥ 2):

r r r r r rp p p1 1 1 1 1 1

v? 1D4:

r r rr?

1

2

1

1

1

Dn (n ≥ 5):

r r r r r rp p pr1 2

1

2 2 2 1

? 1

E6:

r r r r rr?

1 2

2

3 2 1

1 E7:

r r r r r rr?2 3

2

4 3 2 11

E8:

r r r r r r rr?

2 4

3

6 5 4 3 2 1

(8.10)

Let q be a unit form such that ∆(q) is an extended Dynkin diagram. Noticethat the numbers on the vertices of the graphs in (8.10) do not indicate thevertices itself but they define a vector x for which q(x) = 0, as easily can bechecked. Hence q is not positive definite. Moreover, if p is positive definite,then ∆(p) cannot contain an extended Dynkin diagram as subgraph. It isshown in Exercise 8.4.2 that q is positive semidefinite of corank 1 if ∆(q) isan extended Dynkin diagram. ♦

Exercises

8.4.1 Calculate the Euler form of the algebra given in Example 7.8.

8.4.2 Let q : Zn → Z be a unit form whose graph is an extended Dynkin diagram,see list (8.10). Show that q is positive semidefinite of corank 1. Hint: denote byx the vector indicated in (8.10). Verify that x satisfies the following remarkableproperty.

2xi +∑j 6=i

qijxj ,

that is, twice the value xi equals the sum of the values xj over all neighbours j ofi in the graph ∆(q). Use this formula to prove that

q(y + x) = q(y)

for all y ∈ Zn.

Now observe that v? = 1, where ? is a vertex of the extended Dynkin diagram ∆,as indicated in (8.10), such that the restriction to all other vertices is the Dynkindiagram ∆. Now use that every vector w ∈ Zn can be written as w = w′+αv withw′? = 0 and α ∈ Z and conclude that q(w) ≥ 0.

162

8.5 Roots

8.4.3 Let A = KQ/I, where Q =−→A 5 and I is the ideal generated by all compo-

sitions of two consecutive arrows in Q. Calculate the Cartan matrix CA and thegraph ∆(χA).

8.5 Roots

Let q : Zn → Z be a unit form. In general, a vector x ∈ Zn is called a root,or 1-root, of q if q(x) = 1. A root x ∈ Zn is called positive if xi ≥ 0 for alli. A root x ∈ Zn is called negative if −x is positive. We denote set of allroots by

R(q) = x ∈ Zn | q(x) = 1

and the set of all positive roots by

P (q) = x ∈ R(q) | xi ≥ 0 for all i = 1, . . . , n.

By Example 8.4 the canonical basis vectors ei and the dimension vectorsof the indecomposable projective A-modules are roots of χA = qA if Ais a finite-dimensional K-algebra. This shows that there are interestingindecomposable A-modules whose classes in K(A) are roots. But these areno exceptions: by Lemma 8.12

χA(x) = χA(ΦAx).

and therefore [τ−À Px] is a root for each x and ` ≥ 0. Similarly, preinjectiveindecomposables define roots. However not for every indecomposable A-module M the class [M ] is a root of χA, see Exercise 8.5.1.

Proposition 8.19. Let q : Zn → Z be a positive definite quadratic form.Then for any r ∈ Z≥0 the fiber q−1(r) is finite.

Proof. We know from Lemma 8.16 (a) that qR(x) > 0 holds for each x ∈Zn \ 0 and therefore 1

qRis well defined on Rn \ 0. Moreover, since

S1 = x ∈ Rn | ‖x‖ = 1 is compact, qR attains there a global minimum m.

Hence we have for all x with qR(x) = r that r‖x‖2 = qR(x)

‖x‖2 = qR( x‖x‖) ≥ m and

hence ‖x‖ ≤√

rm . This shows that there can be only finitely many vectors

in q−1R (r) ∩ Zn = q−1(r).

Corollary 8.20. If q is a unit form such that its graph ∆(q) is a Dynkindiagram, then there are only finitely many roots of q.

163


Proof. This is an immediate consequence of Proposition 8.19 and Lemma8.17.

Proposition 8.21. Let q be a unit form such that its graph ∆(q) is a Dynkindiagram. Then any root of q is either positive or negative.

Proof. Let v be any root of q. Write x = x+ + x− with x+i ≥ 0, x−i ≤ 0 but

at most one of them non-zero, for each i. Then 1 = q(x) = q(x+) + q(x−) +∑i,j qijx

+j x−i , where the sum is taken over all (i, j) such that x+

i 6= 0 and

x−j 6= 0. Thus the right hand side consists of three non-negative summands,

the latter since qij ≤ 0 and x+i x−j ≤ 0 for any such (i, j). Hence q(x+) = 0

or q(x−) = 0 which implies x+ = 0 or x− = 0, respectively.

If q : Zn → Z is a quadratic form, we define the symmetric bilinear form(−|−)q : Zn × Zn → Z associated to q by

(x|y)q = q(x+ y)− q(x)− q(y). (8.11)

The next rather technical result is very useful in order to determine theroots in practice.

Lemma 8.22. Let q : Zn → Z be a unit form such that ∆(q) is a Dynkindiagram. Then for any positive root x of q with ‖x‖ > 1 there exists somevertex i such that x− ei is again a root.

Proof. It follows from ‖x‖ > 1 that x − ei 6= 0 and therefore q(x − ei) > 0for all indices i. We assume that for all i the vector x−ei is not a root, thatis q(x− ei) > 1 for i = 1, . . . , n.

Since x is a root we have (x|x)q = 2q(x) = 2. On the other hand, it followsfrom

2 ≤ q(x− ei) = q(x) + q(ei)− (x|ei)q = 2− (x|ei)q,

that (x|ei)q ≤ 0. Therefore 2 = (x|x)q =∑

i xi(x|ei)q ≤ 0, a contradiction.

Remark 8.23. The previous lemma is really very practical to determineall positive roots. If x is a positive root, then either ‖x‖ = 1 or ‖x‖ > 1.In the first case we have x = ei for some i. In the second case there existssome i for which x − ei is again a root, and therefore again positive, byProposition 8.21. Hence we can compute all positive roots in the followingway: start from the canonical ones e1, . . . , en and add ei for i = 1, . . . , n to

164

8.5 Roots

each of them and discard those which are not roots. Proceed this way andyou will necessary get all positive roots. ♦

For a given unit form q we define the root diagram Ψ(q) as the graphwhose vertices are the roots of q, that is, Ψ(q)0 = R(q), whereas the edgesare assigned as follows: there is a full edge joining x and y if x+ y or x− yis again a root and a dotted edge if x = −y.

Example 8.24. The following picture shows the root diagram Ψ(q) if∆(q) = A2.

rrrrr r

@@

@@

e1 −e1

e2 −e2

e1 + e2 −(e1 + e2)

♦

Proposition 8.25. Let q : Zn → Z be a unit form such that ∆(q) is aDynkin diagram. Then the root diagram Ψ(q) is connected. Furthermorethe number of roots |R(q)| for the various Dynkin types is as indicated inthe following list.

∆(q)

|R(q)|

An

n(n+ 1)

Dn

2(n− 1)n

E6

72

E7

126

E8

240

As a consequence the number |R(q)| together with n determines ∆(q) unique-ly.

Proof. Clearly, if x is a root, then also −x and x is connected to −x bya dotted edge. Hence, it is sufficient to show that the positive roots areconnected among themselves. First, if i and j are two adjacent vertices inthe graph ∆(q), then clearly ei + ej is a root and hence the roots ei andej are connected. This shows that the roots e1, . . . , en are connected. ByLemma 8.22 and induction over ‖x‖ the connectedness of R(q) follows.

The positive roots of q if ∆(q) = An are of the form 0 p p p01 p p p10 p p p0 wherethe first 1 stands in place ` and the last in place h ≥ `. Clearly, any such

165


vector is a root and furthermore for any such root x the vector x + ei iseither again in the list or not a root as direct inspection shows. Hence thismust be all positive roots and there are n(n+1)

2 of them.

A similar argument shows that the positive roots of q for ∆(q) = Dn are ofthe following form.

00 p p p01 p p p10 p p p00 (n−2)(n−1)2

11 p p p p p p p p p p10 p p p00n− 1 01 p p p p p p p p p p10 p p p01

n− 1

11 p p p p p p p p p p10 p p p01n− 2

12 p p p p p p p p p p21 p p p10 p p p01 (n−3)(n−2)2

We indicated the number of such roots to the right and by adding them upwe get that there are n2−n positive roots and another n2−n negative rootsand therefore 2(n− 1)n roots altogether by Proposition 8.21. The roots forEn with n = 6, 7, 8 can be obtained following Remark 8.23, see Exercise 8.5.3.This shows that |R(q)| has the size as indicated in the statement.

It remains to verify that ∆(q) is uniquely determined by the two numbers nand |R(q)|. But there is only one case when |R(q)| = |R(p)| being ∆(q) and∆(p) two different Dynkin diagrams, namely if ∆(q) = A8 and ∆(p) = E6,but the two unit forms have different number of variables.

Theorem 8.26. In each equivalence class of positive definite unit formsthere exists one and only one unit form q with the property that each con-nected component of ∆(q) is a Dynkin diagram.

Proof. Let q be a positive definite unit form. By Proposition 8.19 the set ofpositive roots P (q) is finite. Furthermore, it follows from

0 ≤ q(ei ± ej) = q(ei) + q(ej)± qij = 2± qij

that −1 ≤ qij ≤ 1 for all i 6= j. If there exists a coefficient qij = 1,then let T = 1n − Eij . Then q′ = qT is again a unit form since q′(ea) =q(ea) = 1 for a 6= j and q′(ej) = q(ej − ei) = 2 − qij = 1. For any x ∈ Zndenote x′ = T−1x. Then q′(x′) = q(x) and we obtain an injective mapP (q) → P (q′), x 7→ x′, since x′h = e>h T

−1x = e>h x = xh for h 6= i andx′i = e>i T

−1x = (ei + ej)>x = xi + xj ≥ 0. However, this inclusion is proper

since ei ∈ N(q′) but Tei 6∈ P (q). So |P (q)| < |P (q′)| ≤ |R(q′)| = |R(q)|.Thus after finitely many steps we must end up with a unit form q(m) such

that q(m)ij ≤ 0 for all i 6= j.

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8.5 Roots

This shows that each positive definite unit form q is equivalent to a unit formp whose graph contains only simple full edges, that is, it does not contain adotted edge nor multiple full edges.

Let ∆ be a component of ∆(p). We want to show that ∆ is contained inthe list (8.8). First of all ∆ cannot contain a cycle because otherwise someAn would be contained. Next, if we look at the degree of a vertex, that is,the number of vertices joined by an edge to it, we see that for each vertexthe degree is 3 or less since otherwise D4 is contained in ∆. Furthermore,∆ has at most one vertex with degree 3 because otherwise ∆ would containDn for some n ≥ 5.

If there is no vertex of degree 3, then we have the case An and if there isone such vertex i, then the lengths l1 ≤ l2 ≤ l3 of the arms (counting thevertices including i) are bounded: using that ∆ does not contain E6, wehave l1 = 2. Using E7 6⊆ ∆, we have l2 = 2, 3 and in case l2 = 3 we useE8 6⊆ ∆ to see that l3 ≤ 5.

It remains to see that in each equivalence class there is at most one such“normal form” p. But the root diagram Ψ(p) is invariant under any equiv-alence and hence the connected components of Ψ(p) indicate the connectedcomponents of ∆(p) and the number of variables involved for a fixed com-ponent is given by the rank of the subgroup generated by the roots of thatcomponent. Hence by Proposition 8.25 the corresponding Dynkin diagramof each component is uniquely determined.

Remark 8.27. The connection between indecomposables and roots can beremarkably strong, see Exercise 8.5.4. That’s why the whole last chapterwill be devoted to it. ♦

Exercises

8.5.1 LetA be the Kronecker algebra. Use Exercise 8.3.2 to show that χA([Rn,λ]) =0 holds for each n ≥ 1 and each λ ∈ K ∪ ∞.

8.5.2 Calculate χA([M ]) for each indecomposable A-module M for A the algebraconsidered in Example 7.8.

8.5.3 Use Remark 8.23 to determine the positive roots of q, where ∆(q) = E8. Usethe list obtained this way to determine the roots for ∆(q) = E6, resp. ∆(q) = E7:show that these are the vectors x in the list which satisfy x7 = x8 = 0, resp. x8 = 0).

8.5.4 There are three quivers whose underlying graph is A3. Knit the preprojectivecomponent in each case and verify directly that always the same dimension vectors

167


occur. Show that these vectors are precisely the positive roots of the unit formq(A3).

168

Chapter 9

Indecomposables anddimensions

Some of the deepest results in the theory of representations of algebras arepresented within this chapter. All of them deal with the representation type,in other words, with the amount of non-isomorphic indecomposable modulesand their dimension vectors.

9.1 The Brauer-Thrall conjectures

In this last chapter we gathered some results concerning the representationtype and in particular dimensions and dimension vectors of indecomposablemodules. We will not be able to proof all of them, for many we have torefer to the literature. We start with two early statements, brought intolife by Brauer around 1940 as exercises for his students and which remainedunproved for a long time. Nowadays they are known as the first and secondBrauer-Thrall conjecture. Both statements concern the representationtype of an algebra. We prove the first Brauer-Thrall conjecture and refer tothe literature for the second since its proof is well beyond the scope of thisbook.

For a fixed, finite-dimensional K-algebra A with quiver Q we denote byκ(m) ∈ N ∪ ∞ the number of isomorphism classes of indecomposableA-modules with dimension m. The algebra A is called of bounded rep-resentation type if there exists some number M > 0 such that κ(m) = 0for all m ≥ M . If A is not of bounded representation type, it is called of

169

9 Indecomposables and dimensions

unbounded representation type. The algebra A is called of stronglyunbounded representation type if there exists an infinite sequence ofdimensions m0 < m1 < m2 < . . . such that κ(mi) =∞ for all i ∈ N.

For the proof of the first Brauer-Thrall conjecture we need the followingtechnical result.

Lemma 9.1. Let A is a finite-dimensional algebra. Furthermore letM0, . . . ,Mt be indecomposable A-modules and

M0f1−→M1

f2−→M2 → . . .→Mt−1ft−→Mt

be radical homomorphisms. If t ≥ 2d − 1 and dimMi ≤ d for all 0 ≤ i ≤ t,then the composition ft · · · f1 is zero.

Proof. The crucial part in the proof is to show the following claim: If g andh are homomorphisms

Lg−→Mi−1

fi−→Mih−→ N

such that dimK Im(g) ≤ ` and dimK Im(h) ≤ ` for some ` > 0, then itfollows that dimK Im(hfig) ≤ ` − 1. Assume the contrary. Then, sinceIm(hfig) ⊆ Im(hfi) ⊆ Im(h) we have

` ≤ dimK Im(hfig) ≤ dimK Im(hfi) ≤ dimK Im(h) ≤ `

and therefore all these dimension must be equal to `. This shows thatKer(hfi)∩Im(g) = 0 and therefore Ker(hfi)⊕Im(g) is a submodule of Mi−1.Since Im(hfi) ' Mi−1/Ker(hfi) we get that dimK Ker(hfi) = dimMi−1 −dimK Im(hfi) = dimMi−1 − `. Hence

dimK (Ker(hfi)⊕ Im(g)) = dimKMi−1 − `+ ` = dimKMi−1

showing that Mi−1 = Ker(hfi) ⊕ Im(g). Since Im(g) 6= 0 and Mi−1 isindecomposable we conclude that Ker(hfi) = 0. This shows that hfi andhence fi is injective. Similarly, we conclude from

` ≤ dimK Im(hfig) ≤ dimK Im(fig) ≤ dimK Im(g) ≤ `,

that all these dimensions equal ` and therefore Ker(h) ∩ Im(fig) = 0. SinceIm(h) ' Mi/Ker(h) we get dimK Ker(h) = dimKMi − ` and thereforedimK (Ker(h)⊕ Im(fig)) = dimMi. This shows that Mi = Ker(h)⊕Im(fig)and since fig 6= 0 and Mi is indecomposable we conclude that Ker(h) = 0

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9.1 The Brauer-Thrall conjectures

and Mi = Im(fig). Consequently fig and hence fi is surjective. Hence fi isbijective in contradiction to the assumptions. This shows the claim.

Now, if we take g = idMi−1 and h = idMi , then dimK Im(g) ≤ d anddimK Im(h) ≤ d. Thus, we conclude in the first step that dimK Im(fi) ≤d−1 for all i. In the second step we set g = fi−1 and h = fi−1 and concludethat the image fi+1fifi−1 has dimension at most d− 2 for all i. Inductivelywe get that

dimK Im(fi+m · · · fi) ≤ d− p

if m ≥ 2p − 1. The result follows then by setting p = d.

Theorem 9.2 (First Brauer-Thrall conjecture). An algebra of bounded rep-resentation type is of finite representation type.

Proof. Suppose A is of bounded representation type. Then there exists somenumber d > 0 such that κ(m) = 0 for all m > d. In other words, dimKM ≤d for each indecomposable A-module M . We now will define a series of finitesubsets of the Auslander-Reiten quiver ΓA of A. The first set consists of allindecomposable projective A-modules S0 = P1, . . . , PN ⊆ ΓA. Inductivelyif Si is defined, we define

Si+1 = Y ∈ ΓA | there exists an arrow X → Y in ΓA with X ∈ Si

which, since Si is finite, is again finite by Remark 7.3. Note that by con-struction, each radical morphism X → Y with X ∈ Si factors over a directsum of A-modules in Si+1.

Thus, if M is an indecomposable A-module then there exists a non-zeromorphism f : P →M for some indecomposable projective A-module P ∈ S0,then either f is an isomorphism andM belongs to S0 or the factorization overS1 yields two homomorphisms f1 : P →M1 and g1 : M1 →M with g1f1 6= 0with f1 radical and M1 ∈ S1. In the next step we see that if g1 is notan isomorphism, the factorization over S2 yields a radical homomorphismf2 : M1 → M2 ∈ S2 and g2 : M2 → M with g2f2f1 6= 0. Inductively we

obtain that either M belongs to⋃2d−1m=0 Sm or there exist homomorphisms

Pf1−→M1

f2−→M2 → . . .→M2d−2

f2d−1−−−→M2d−1

g2d−1−−−→M

with g2d−1f2d−1 · · · f2f1 6= 0. Since the latter is impossible by Lemma 9.1,we obtain that each indecomposable A-module belongs to the finite set⋃2d−1m=0 Sm.

171


The first Brauer-Thrall conjecture states that it is not possible to haveκ(m) = ∞ for some small dimensions m but κ(m) = 0 for all large m.In other words: a finite-dimensional K-algebra of infinite representationtype has indecomposable modules of arbitrarily large dimension. The firstBrauer-Thrall conjecture was first proved by Roiter in [23] for algebraicallyclosed fields and later generalized by Auslander in [1].

Theorem 9.3 (Second Brauer-Thrall conjecture). An algebra of unboundedrepresentation type is of strongly unbounded representation type.

This result states that if an algebra is of infinite representation type, thenthere exists infinitely many dimensions for which there are infinite familiesof indecomposable modules. The first complete proof of the second Brauer-Thrall conjecture was given in [3] by Bautista for algebraically closed fieldsof characteristic different from 2. For arbitrary fields, the problem remainsopen, see also Ringel’s report [21] for a detailed discussion about the devel-opments in the proofs of the Brauer-Thrall conjectures.

9.2 Gabriel’s Theorem

Gabriel’s Theorem is the first classification result in the representation the-ory of finite-dimensional algebras, which successfully links the indecompos-able representations to the roots of the Tits form (8.5) for a special classof algebras. It classifies those hereditary algebras which are of finite rep-resentation type. First we prove an auxiliary result concerning short exactsequences.

Lemma 9.4. Let A be a finite-dimensional K-algebra and let

ε : 0→Ma−→ E

b−→ N → 0

be a short exact sequence of A-modules. If the sequence ε is not split, thenthe following inequality holds:

dimK EndA(E) < dimK EndA(M ⊕N).

Proof. Applying the functor HomA(−, E) to the exact sequence ε we get

0→ HomA(N,E)HomA(b,E)−−−−−−−→ EndA(E)

HomA(a,E)−−−−−−−→ HomA(M,E)

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9.2 Gabriel’s Theorem

and therefore

dimK EndA(E) = dimK HomA(N,E) + dimK(Im HomA(a,E))

≤ dimK HomA(N,E) + dimK HomA(M,E).(9.1)

With a similar reasoning we get by applying HomA(M,−) to ε the followinginequality

dimK HomA(M,E) ≤ dimK HomA(M,M) + dimK HomA(M,N)

= dimK HomA(M,M ⊕N).(9.2)

Finally, by applying HomA(E,−) to ε, we get the exact sequence

0→ HomA(N,M)→ HomA(N,E)→ HomA(N,N)δ 6=0−−→ Ext1

A(N,M),

where δ is the connecting homomorphism, which is non-zero since δ(idN ) =ε 6= 0. Therefore we get a strict inequality

dimK HomA(N,E) < dimK HomA(N,M) + dimK HomA(N,N)

= dimK HomA(N,M ⊕N).(9.3)

Substituting (9.2) and (9.3) in (9.1) yields

dimK EndA(E) < dimK HomA(M,M ⊕N) + dimK HomA(N,M ⊕N)

= dimK EndA(M ⊕N)

which is what we wanted to prove.

Theorem 9.5 (Gabriel). Let A = KQ be the path algebra of a quiver Qwithout cycle. Denote by qA the Euler form of A. Then A is of finiterepresentation type if and only if qA is positive definite or equivalently ifand only if Q is the disjoint union of connected quivers whose underlyinggraphs are Dynkin diagrams. Moreover, in that case, the function

Ψ: (ΓA)0 → P (qA), X 7→ dimX

is bijective, where P (qA) denotes the set of positive roots of qA.

Proof. By Remark 8.14 the Euler form and the Tits form coincide, χA = qA,since A is hereditary. Note that it is enough to prove the result for connectedA. Otherwise A = A1×A2 and qA = qA1 × qA2 holds and then A is of finiterepresentation type if and only if A1 and A2 both are so and similarly qA ispositive definite if and only if qA1 and qA2 are both positive definite.

173


Next we observe that by Proposition 7.10 there exists a unique preprojec-tive component C in ΓA since A is hereditary and connected. Hence weconclude from Theorem 7.7 that A is of finite representation type if andonly if C is finite and that this happens if and only if C = ΓA. Further-more, it follows from Proposition 7.11 that each A-module M ∈ C satisfiesdimK EndA(M) = 1 and Ext1

A(M,M) = 0. Therefore dimM is a root of qAfor each M ∈ C.

Now suppose that qA is positive definite. Then it follows from Proposi-tion 8.19 that there are only finitely many roots of qA. Therefore, if wecan show that for each root d there exists at most one indecomposable A-module M ∈ C such that dimM = d, then we have shown that C is finite.So, assume that M,N ∈ C are two indecomposable A-modules with thesame dimension vector d. Then

1 = qA(d) = 〈dimM,dimN〉 = dimK HomA(M,N)− dimK Ext1A(M,N)

shows that HomA(M,N) 6= 0. Similarly, we get HomA(N,M) 6= 0 andconclude that M = N since there is no cycle in C.

Conversely assume now that C is finite. Note that, if we can show that foreach positive root d of qA there exists at least one indecomposable A-moduleM with dimM = d, then there exist only finitely many roots since ΓA = C.So, assume that d is a positive root of qA. Since d is positive it is possible tofind A-modules with dimM = d, for example M =

⊕i∈Q0

Sdii has dimensionvector d. Choose such an A-module M such that dimK EndA(M) is minimal.Decompose M into indecomposables M =

⊕ti=1Mi. By Proposition 7.11 we

have Ext1A(Mi,Mi) = 0 and it follows from Lemma 9.4 that Ext1

A(Mi,M′i) =

0, where M ′i =⊕

j 6=iMj , since otherwise there would exist some non-splitshort exact sequence M ′i → E →Mi and consequently

dimK EndA(E) < dimK EndA(Mi ⊕M ′i) = dimK EndA(M),

a contradiction to the minimality of dimK End(M) since dimE = dimMi +dimM ′i = dimM . This shows that Ext1

A(M,M) = 0. Then it follows from

1 = qA(dimM) = dimK EndA(M) ≥t∑i=1

dimK EndA(Mi) = t

that t = 1, in other words, that M is indecomposable.

Observe that we also proved the claimed bijection between the indecompos-able A-modules and the positive roots of qA.

174

9.3 Reflection functors

Theorem 9.5 was first proved by Gabriel in [10]. Soon a more conceptualproof was given by Bernstein, Gelfand and Ponomarev in [4]. The presentedapproach is taken from Ringel [22].

Exercises

9.2.1 Generalize Theorem 9.5 to finite-imensional K-algebras A which have apreprojective component C in the Auslander-Reiten quiver ΓA and which satisfygldimA ≤ 2 by adjusting the proof slightly to the more general setting.


Observe that Gabriel’s Theorem 9.5 implies that the dimension vectors ofthe indecomposables do not depend on the orientation of the quiver, anunexpected fact, which already could be observed in Exercise 8.5.4. In thisand the next section we focus more closely on this peculiarity of hereditaryalgebras.

Let Q be a finite quiver without cycle. Note that this implies that Q alwayshas a source and a sink, see Exercise 9.3.1.

If the x is a source or a sink of Q, we denote by σxQ the quiver which isobtained from Q by reversing the direction of all arrows which start or endin x. Thus if x is a source of Q, then it will be a sink in σxQ and vice versa.We now define functors σx : repQ→ repσxQ in the following way. Let firstx be a source of Q. Then for each representation V ∈ repQ we have a linearmap

Vx→• : Vx

...Vα...

−−−→

⊕α∈Q1 : s(α)=x

Vt(α)

and denote its cokernel by⊕α∈Q1 : s(α)=x

Vt(α)

[· · · V ′α · · ·]−−−−−−−→ V ′x.

Hence we get a representation σxV of σxQ which is defined by

(σxV )y =

Vy, if y 6= x,

V ′x, if y = x,(σxV )β =

Vβ, if t(β) 6= x,

V ′β, if t(β) = x.

175


Now if f : V → W is a morphism of representations of Q, then we definef ′x : V ′x → W ′x as the map induced by

⊕α fα :

⊕α Vt(α) →

⊕αWt(α). This

defines the functor σx if x is a source of Q and the inverse constructionworks if x is a sink of Q. These functors were called Coxeter reflectionfunctors by Bernstein, Gelfand and Ponomarev in [4] and are nowadaysalso called BGP-reflection functors.

Proposition 9.6. Let Q be a finite quiver without cycle and x a source ora sink of Q. Then σxSx = 0 and if V is an indecomposable representationof Q satisfying V 6' Sx, then σxV is an indecomposable representation ofσxQ satisfying σxV 6' Sx. In particular σx induces a bijection between theisomorphism classes of the indecomposable representations V 6' Sx of Q andthe isomorphism classes of the indecomposable representations W 6' Sx ofσxQ.

Proof. Notice first that for all representations V we have V ' Sdx ⊕ Wwhere d is the dimension of KerVx→•. Hence, if V is indecomposable, theneither V = Sx or Vx→• is injective. It follows directly from the definitionthat σxSx = 0 and that Sx cannot be a direct summand of σxV for anyrepresentation V of Q. If Vx→• is injective then σxσxV ' V . Now, ifV = U ⊕W , then σxV = σxU ⊕ σxW . Thus if V is indecomposable andV 6' Sx, then σxV cannot be decomposable since otherwise V = σxσxVwould also be decomposable. This finishes the proof.

Exercise 9.3.2 gives an even stronger version of Proposition 9.6. We nowstudy how the reflection functors change the dimension vectors.

Lemma 9.7. Let Q be a finite quiver without cycle, x a source of Q and qthe quadratic form of the path algebra KQ. Then for any representation Vof Q with injective map Vx→• the dimension vector of σxV is given by

dimσxV = dimV − (dimV |ex)qex,

where (x|y)q = q(x+y)−q(x)−q(y) is the symmetric bilinear form associatedto q.

Proof. Since Vx→• is injective we have a short exact sequence

Vx →⊕

α∈Q1 : s(α)=x

Vt(α) → V ′x

176


and therefore we can calculate dimσxV as follows:

dimσxV = dimV + δex,

where δ =∑

α∈Q1 : s(α)=x dimK Vt(α) − 2 dimK Vx. It follows from the de-scription of the Tits form given in (8.6) that for v = dimV we have

(v|ex)q = q(v + ex)− q(v)− q(ex)

=∑i∈Q0

(v + ex)2i −

∑(i→j)∈Q1

(v + ex)i(v + ex)j

−∑i∈Q0

v2i +

∑(i→j)∈Q1

vivj − 1.

After cancelling all possible terms we are left with (v|ex)q = 2vx−∑

i→x vi−∑x→j vi and since x is a source we get

∑i→x vi = 0 and therefore conclude

that (v|ex)q = δ.

We define the linear maps

sx : K(KQ)→ K(KQ), sx(v) = v − (v|ex)qex, (9.4)

and get that the dimension vectors of representations change under σx ac-cording to sx in the sense that

dimσxV = sx(dimV ), (9.5)

if the representation V does not have any direct summands isomorphic toSx. In the following we assume that the vertices of Q are 1, . . . , n andtherefore K(KQ) ' Zn. We extend the symmetric bilinear form (−|−)q toRn × Rn → Rn and the linear maps sx to Rn → Rn in the obvious way.

Lemma 9.8. With the above definitions the following properties hold.

(a) sx is an idempotent, that is, s2x = idRn,

(b) sx is orthogonal, that is, (sxv|sxw)q = (v|w)q,

(c) sx reflects ex, that is, sx(ex) = −ex,

(d) Rn = Rex ⊕ e⊥x where e⊥x = v ∈ Rn | (v|ex) = 0.

177


Proof. Note that (ex|ex)q = q(2ex) − 2q(ex) = 2 and hence (c) follows im-mediatly. All other properties follow from these two facts. Since

s2x(v) = sx(v)− (v|ex)qsx(ex) = v − (v|ex)qex + (v|ex)qex = v.

property (a) follows. Also (b) is straightforward:

(sxv|sxw)q = (v − (v|ex)qex | w − (w|ex)qex)

= (v|w)q − (v|ex)q(ex|w)q

− (w|ex)q(v|ex)q + (v|ex)q(w|ex)q(ex|ex)q

= (v|w)q

To see (d) write any v ∈ Rn as

v =1

2(v|ex)qex +

(v − 1

2(v|ex)qex

)and observe that the second summand lies in e⊥x since (v − 1

2(v|ex)qex |ex) = (v|ex)q − 1

2(v|ex)q(ex|ex)q = 0. This shows that Rex + e⊥x = Rn. Since(ex|ex)q 6= 0 we also have Rex ∩ e⊥x = 0 showing the result.

Remarks 9.9 (a) It is really formula (9.5) which gave the idea of namingthe functors σx Coxeter reflections since this formula is well known fromthe study of reflection groups and the generalization to Coxeter groups, see[12] for an elementary introduction. It also justifies to call the maps sxreflections even when the bilinear form (−|−)q is not positive definite.

(b) Bernstein, Gelfand and Ponomarev used in [4] this reflection functorsto give a proof of the bijection statement in Gabriel’s Theorem. It is possibleto reach any root of a positive definite unit form by starting from the rootse1, . . . , en and using these reflections, see Exercise 9.3.3.

We would like to use this for representations: It is clear that for each jthere exists, up to isomorphism, a unique indecomposable representationwith dimension vector ej , namely the simple representation Sj . So letv = sit · · · si2si1(ej) be a positive root obtained by reflections. Now, ifi1 is a source or a sink of Q and iteratively ih is a source or a sink of thequiver σih−1

· · ·σi1Q for 1 < h ≤ t, then we see that there exists a unique in-decomposable representation with dimension vector v, namely σit · · ·σi1Sj .Unfortunately if for some h the vertex ih is not a source or a sink of thequiver σih−1

· · ·σi1Q then this construction does not work. For the case ofDynkin diagrams this can be fixed, see [4]. ♦

178

9.4 Kac’s Theorem

The next section shows the most general result concerning the dimensionvectors of indecomposable representations obtained so far.

Exercises

9.3.1 Prove that a finite quiver Q without cycle always has at least one sourceand one sink. Show that both hypothesis are necessary by giving two examples ofquivers without source nor sink: one which is finite but has some cycle and anotherwhich is infinite but without cycle.

9.3.2 Let x be a source or a sink of a finite quiver Q. Prove that the Coxeter reflec-tion functors induce equivalences between the categories repQ/Ix and repσxQ/Jxwhere Ix (resp. Jx) is the ideal of repQ (resp. of repσxQ) consisting of all mor-phisms which factor over Sdx for some d > 0.

9.3.3 Let Q be a Dynkin quiver, that is, a quiver whose underlying graph is aDynkin diagram. Then the associated unit form q = qKQ is poisitive definite andby Lemma 8.22 for each root v with ‖v‖ > 1 there exists an index i such that v−eiis again a root. Show that in this case v − ei = sx(v). This shows that in the caseof Dynkin diagrams it is possible to reach any root using the reflections s1, . . . , snstarting from the roots e1, . . . , en.

9.4 Kac’s Theorem

Let Q be a finite quiver with vertices 1, . . . , n and let q be the associatedquadratic form. The group W generated by the reflections s1, . . . , sn is calledthe Weyl group. The Weyl group acts on Zn in the obvious way.

In case Q is a Dynkin quiver, we know from Theorem 9.5 that the positiveroots of q correspond to the indecomposable representations of Q. In viewof Exercsie 9.3.3 we can say that the positive roots of q are precisely thosepositive vectors which can be obtained by a sequence of reflections sx asdefined in (9.4). In other words, the set P (q) of positive roots equals Nn ∩We1, . . . , en.It is remarkable that this relationship extends to all finite quivers withoutcycle if the notion of roots is enhanced the right way. To do so, we definethe support supp v of a vector v ∈ Zn as the full subquiver of Q given bythose vertices i for which vi 6= 0. Furthermore, the set

F = v ∈ Nn | supp v is connected and (v|ej)q ≤ 0 for 1 ≤ j ≤ n (9.6)

is called the fundamental region, where (−|−)q is the symmetric bilinearform (8.11) associated to q.

179


A vector v ∈ Zn is called a Schur root if it belongs to the W -orbit WS,where S = e1, . . . , en ∪ F ∪ (−F ). The elements of the form wej for somej = 1, . . . , n and some w ∈ W are called real whereas the elements of theform w(±f) for some f ∈ F and some w ∈ W are called imaginary. Notethat q(v) = 1 for any real Schur root v since by Lemma 9.8(b) the Weylgroup preserves the bilinear form (−|−). Similarly one can see that q(v) ≤ 0holds for any imaginary Schur root, see Exercise 9.4.1. Moreover, if v is areal Schur root, then ±v are the unique multiples of v which are Schur roots,whereas if v is imaginary, then every non-zero multiple of v is a Schur root,see Exercise 9.4.2.

Theorem 9.10 (Kac). Let K be any field and Q a finite quiver withoutcycle. If V is an indecomposable representation of Q, then dimV is a Schurroot. Conversely we have the following characterization.

(a) If v is a positive real Schur root, then up to isomorphism there existsa unique indecomposable representation V of Q with dimV = v.

(b) If v is a positive imaginary Schur root, then there exist infinitely manypairwise non-isomorphic indecomposable representations V of Q withdimV = v.

The proof of Kac’s Theorem is too difficult to give in this notes. It wasproved by Kac in [13] and later generalized by Kac in [14] to quivers withloops. For an approach to Kac’s Theorem using algebraic geometry (asbriefly outlined in Section 9.6) we refer to Kraft and Riedtmann [17].

Remark 9.11. Note that it is a premature idea to think that by Kac’sTheorem we already gain access to all finite-dimensional algebras using thefact that modA is equivalent to a full subcategory of repQ, where Q is thequiver of A. If A is Morita equivalent to KQ/I for some admissible idealI, then Kac’s Theorem yields those dimension vectors for which indecom-posable A-modules may exist, but since we have no control of how manyof those representations satisfy the ideal I we cannot deduce more than thefirst statement: if M is an indecomposable A-module, then dimM is a Schurroot. ♦

Exercises

9.4.1 Let q be the quadratic form associated to a quiver Q. Prove that q(f) ≤ 0for any f in the fundamental region F given in (9.6).

180

9.5 Tame and wild

9.4.2 A vector v ∈ Zn is called indivisible if v = λw for w ∈ Zn and λ ∈ Zimplies that λ = ±1. Show that if v is indivisible, then also sx(v) is indivisible andconclude from this that each real Schur root is indivisible. Prove further that ifv ∈ Zn is an imaginary Schur root then λv is also an imaginary Schur root for eachλ ∈ Z \ 0.

9.5 Tame and wild

At the end of Chapter 1 we found three very distinct phenomena in classify-ing indecomposable elements of matrix problems. We later translated theseproblems into the language of representations of a quiver and modules overan algebra. We now come back to these three types of complexity and studythem in more detail. We assume throughout the rest of this chapter that Kis algebraically closed.

To compare the complexity of the classification problems for two differentfinite-dimensional K-algebras A and B, we introduce the following language.A functor F : modA → modB preserves indecomposability if it mapsindecomposable A-modules to indecomposable B-modules and F reflectsisomorphisms if FM ' FN implies M ' N for any A-modules M and N .Thus, if we have a functor F : modA → modB which preserves indecom-posability and reflects isomorphisms, then a classification of indecomposableB-modules would yield a classification of indecomposable A-modules. How-ever, note that such a functor does not necessarily transport a lot of structurefrom modA to modB, see Exercsie 9.5.1.

We start by studying the wild behaviour in some detail. Let Q(t) be thequiver with one vertex 0 and t loops α1, . . . , αt. A representation of Q(t) isa vector space together with t (not necessarily commuting) endomorphisms.Such a representation it is given by t square matrices (Vα1 , . . . , Vαt). Twosuch tuples (Vα1 , . . . , Vαt) and (Wα1 , . . . ,Wαt) represent isomorphic repre-sentations if and only if they are simultaneously conjugate, that is, if thereexists an invertible square matrix U such that Wαi = UVαiU

−1 for alli = 1, . . . , t.

Lemma 9.12. There exists a functor F : repQ(t) → repQ(2) which pre-serves indecomposability and reflects isomorphisms.

Proof. Given a representation V ∈ repQ(t) we define FV by setting (FV )0 =

181


(V0)t+1 and

(FV )α1 =

0 id 0 · · · 00 0 id · · · 0...

......

. . ....

0 0 0 · · · id0 0 0 · · · 0

, (FV )α2 =

0 Vα1 0 · · · 00 0 Vα2

· · · 0...

......

. . ....

0 0 0 · · · Vαt

0 0 0 · · · 0

.

If V and W are representations of Q(t) then a morphism g : V → W isgiven by single linear map g = g0 : V0 → W0 such that Wαig = gVαi fori = 1, . . . , t. To define Fg : FV → FW we observe first that any morphismh : FV → FW is given by a matrix of size (t+ 1)× (t+ 1) whose entries hijare morphisms V0 → W0 satisfying (FW )α`h = h(FV )α` for ` = 1, 2. Thuswe see that

Fg =

g 0 0 · · · 00 g 0 · · · 00 0 g · · · 0...

......

...0 0 0 · · · g

defines a morphism FG : FV → FW of representations of Q(2). By defini-tion F is functorial and faithful, that is, Fg 6= 0 for g 6= 0. Now supposethat h : FV → FW is a morphism. Then it follows from (FW )α1h =h(FV )α1 that h11 = h22 = . . . = htt = ht+1,t+1 and hij = 0 for i > j. Usingthis fact we conclude from (FW )α2h = h(FV )α2 that Wαih11 = h11Vαi forall i = 1, . . . , t, that is, h11 is a morphism V →W . Thus, if h is an isomor-phism, then so must be h11 since h = (hij) is upper triangular. This showsthat F reflects isomorphisms.

Furthermore, if FV decomposes into two non-zero summands, then thereexist two non-zero idempotents h, h′ ∈ EndQ(2) such that h+h′ = idFV andhh′ = h′h = 0. So h11 and h′11 are two non-zero idempotents in EndQ(t) withh11 + h′11 = idV and h11h

′11 = h′11h11 = 0 showing that V is decomposable.

Thus F preserves indecomposability.

A functor F : C → D is called full if for every pair of objects x, y ∈ C wehave FC (x, y) = D(Fx, Fy) and the functor is called an embedding if itis full and faithful.

Lemma 9.13. Let A and B be two finite-dimensional K-algebras. Anyembedding modA → modB preserves indecomposability and reflects iso-morphisms.

182

9.5 Tame and wild

Proof. Let F : modA → modB be a full functor and M , N be two A-modules. If ϕ : FM → FN is an isomorphism, then ϕ = Fψ for someψ : M → N and also ϕ−1 = Fη for some η : N → M . Now, F (idM ) =idFM = ϕ−1ϕ = FηFψ = F (ηψ) implies idM = ηψ. Similarly, we getidN = ψη and therefore M ' N . This implies that F reflects isomorphisms.

As a consequence a non-zero A-module cannot be mapped to zero. Now, ifFMdecomposes, say FM ' Q′ ⊕ Q′′ with Q′ 6= 0 and Q′′ 6= 0, then thereexist non-zero, orthogonal idempotents ϕ′, ϕ′′ : FM → FM with ϕ′ + ϕ′′ =idFM . Since F is full there exist ψ′, ψ′′ such that ϕ′ = Fψ′ and ϕ′′ =Fψ′′. Observe that ψ′, ψ′′ are non-zero, orthogonal idempotents satisfyingψ′ + ψ′′ = idM . This implies that M ' Imψ′ ⊕ Imψ′′ with Imψ′ 6= 0 andImψ′′ 6= 0. Therefore F preserves indecomposability.

The next result resumes that solving the classification problem for Q(2)

would solve the classification problem for all finite-dimensional K-algebrasat once.

Proposition 9.14. For each finite-dimensional algebra A there exists afunctor modA → repQ(2) which preserves indecomposability and reflectsisomorphisms.

Proof. Since K is algebraically closed Gabriel’s Theorem 3.28 implies thatthere exists a quiver Q and an admissible ideal I of KQ such that modA isequivalent to repI Q. Now, the inclusion J : repI Q → repQ is an embed-ding and therefore preserves indecomposability and reflects isomorphismsby Lemma 9.13.

Let t be the number of arrows of Q, say Q1 = β1, . . . , βt. Then we definea functor G : repQ→ repQ(t) in the following way. For a representation Vof Q we define the representation GV of Q(t) by (GV )0 =

⊕x∈Q0

Vx and

(GV )αh = ιVt(βh)VβhπVs(βh), that is, the composition of the following maps

⊕x∈Q0

VxπVs(βh)−−−−→ Vs(βh)

Vβh−−→ Vt(βh)

ιVt(βh)−−−→

⊕x∈Q0

Vx.

If ψ : V → W is a morphism of representations of Q then we define themorphism Gψ : GV → GW by Gψ =

⊕x∈Q0

ψx. Thus by definition G isfaithful.

To see that G is full, we assume f : GV → GW to be a morphism of repre-sentations of Q(t). Then for each x ∈ Q0 we define ϕx = πWx fι

Vx : Vx →Wx.

183


We want to show that the family ϕ = (ϕx)x∈Q0 is a morphism of represen-tations V →W . For this assume that β = βh : x→ y is an arrow. Then weget that f(GV )αh = (GW )αhf , that is,

fιVy VβπVx = ιWy Wβπ

Wx f. (9.7)

Applying πWy ? ιVx to equation (9.7) we get

πWy fιVy Vβ = Wβπ

Wx fι

Vx

hence ϕyVβ = Wβϕx. This shows that ϕ is indeed a morphism of represen-tations and f = Gϕ. Thus G is a full functor and therefore by Lemma 9.13preserves indecomposability and reflects isomorphisms.

By Lemma 9.12 there exists a functor F : repQ(t) → repQ(2) which pre-serves indecomposability and reflects isomorphisms, so the composition

modA ' repI QJ−→ repQ

G−→ repQ(t) F−→ repQ(2)

preserves indecomposability and reflects isomorphisms. This finishes theproof.

A finite-dimensional K-algebra A is called wild if there exists a functorF : repQ(2) → modA which preserves indecomposability and reflects iso-morphisms. Thus a full classification of all modules of some wild algebrawould yield the classification for all finite-dimensional K-algebras at once.This is considered a “hopeless situation”.

Now let us look at the phenomenon of tameness. Since the formal definitionof tameness is more involved than that of wildness we will approach thedefinition gradually and start with an example of a tame algebra, where wehave the full knowledge about its indecomposables.

Example 9.15. Let A be the Kronecker algebra. We know that the inde-composables occur in every dimension. For odd m ∈ N there exist preciselytwo indecomposable A-modules, namely Q` and J` where 2` + 1 = m. Foreven m ∈ N, say m = 2`, there exist infinitely many indecomposables,namely R`,λ where λ ∈ K ∪ ∞.Thus for every even dimension m = 2` there exists a one-paremeter family.Rephrasing it differently, we can obtain each of these indecomposables, ex-cept Rn,∞, by specializing the indeterminate X to λ in the following infinite-dimensional representation of the Kronecker quiver:

184

9.5 Tame and wild

K[X]` K[X]`,--

J(`,X)

1`

where J(`,X) is the matrix in K[X]`×` with entries

J(`,X)ij =

X, if i = j

1K , if i = j − 1

0, else.

If we specialize X to λ then any polynomial in X becomes an element of Kand thus the polynomial algebra K[X] “specializes” to K. In the following,we will see how to formalize this properly. ♦

Fix a finite-dimensional K-algebra A. Of course a “one-paremeter family ofA-modules” (Mλ)λ should not mean any arbitrary assignment λ 7→Mλ sincewe do not seek merely to index the modules in the family. First of all, eachconsidered modules should have the same dimension vector d ∈ NQ0 . Oneway to visualize something which comes close to a one-paremeter family isa fixed family (Vx)x∈Q0 of vector spaces in the vertices of the quiver Q ofA and a family of matrices (Vα,λ)α∈Q1 such that for each α the family ofmatrices Vα,λ depends polynomially on λ.

Instead of thinking Vα,λ as depending polynomially on λ we can think of asingle matrix Vα whose entries belong to the polynomial algebra K[X], thatis, a matrix in K[X]dy×dx , if α : x → y and d ∈ NQ0 denotes the dimensionvector of the family. To recover a single element of the family we onlyhave to specialize X → λ. Thus we can resume: to give a one-parameterfamily of representations of the quiver Q of A we give a representation inthe ring K[X]. Recalling the equivalence between repI Q and modKQ/I wesee that such a family corresponds to an A-module whose underlying vectorspace is V =

⊕x∈Q0

K[X]dx and whose multiplication with A consists of

multiplications with matrices in K[X]m×m, where m =∑

x∈Q0dx.

Hence V becomes a K[X]-module which is finitely generated free. Sincethe multiplication of A is compatible with the multiplication of K[X] weget an A-K[X]-bimodule V which is finitely generated free as right K[X]-module. We have reached the final definition: a one-parameter family ofA-modules is an A-K[X]-bimodule which is finitely generated free as rightK[X]-module.

The recovering of the A-module Mλ from V formally is done by tensoring

185


with K[X]/(X− λ), that is

Mλ = V ⊗K[X] K[X]/(X− λ).

We are now able to give a formal definition of tameness. A finite-dimensionalK-algebra is called tame if for each dimension vector d there exists a finitefamily FLdi=1 of one-paremeter families such that the following two propertieshold.

(i) Up to isomorphism, there exist only finitely many indecomposable A-modules with dimension vector d which are not isomorphic to Fi,λ =Fi ⊗K[X] K[X]/(X− λ), for some i and some λ.

(ii) Up to finitely many exceptions, the A-modules Fi,λ are indecomposableand pairwise non-isomorphic.

Example 9.16. Each algebra of finite representation type is tame: justspecify the empty family for each dimension vector. Then (ii) is empty and(i) holds since we can allow finitely many exceptions. ♦

The crucial result concerning the classes of tame and wild algebras is thefollowing.

Theorem 9.17 (Tame-Wild Dichotomy). Each finite-dimensional algebraover an algebraically closed field is either tame or wild.

The proof is too involved and we hence have to refer to the literature: Thefirst proof of Theorem 9.17 was given in [9] by Drozd in 1980. Crawley-Boevey rewrote and cleaned it in [6]. In 1993 Gabriel, Nazarova, Roiter,Sergeijuk and Vossieck gave an alternative proof, see [11], improving theresult as mentioned above.

Let A = KQ/I be a finite-dimensional K-algebra given by a quiver Q andan admissible ideal I of KQ. We assume that A admits a “two-parameterfamily of indecomposables”, that is, there exists a dimension vector d ∈NQ0 such that in repI(Q, d) there exists a two-parameter family of moduleswhich are (except finitely many exceptions) indecomposable and pairwisenon-isomorphic. We observe that then A can not be tame and hence mustbe wild by the Tame-Wild Dichotomy Theorem 9.17. In that case, for anyt ≥ 1 there exists a dimension vector d such that repI(Q, d) contains at-parameter family of indecomposable pairwise non-isomorphic A-modules.

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9.6 Module varieties

Exercises

9.5.1 Let A be a finite-dimensional K-alegbra. Show that the canonical projectionfunctor modA → modA/ radA preserves indecomposablity and reflects isomor-phisms.

9.5.2 Show that there exists a functor F : repQ(2) → K[X,Y] which preserves in-decomposablity and reflects isomorphisms. Hint: define the ideal I = 〈α1α2−α2α1〉of the infinite dimensional K-algebra KQ(2) and show that K[X,Y] ' KQ(2)/I.For a representation V of Q(2) define the representation FV of Q(2) by setting(FV )0 = V 3

0 and

(FV )α`=

0 idV0Vα`

0 0 idV0

0 0 0

for ` = 1, 2. Show that FV satisfies the ideal I and define F on morphisms to geta functor F : repQ(t) → repI Q

(t) and then verify that F satisfies the two claimedproperties.


In this section we treat a geometric approach to the question about thedimension vectors of indecomposables and the number of indecomposablesof fixed dimension vector. We give here only a quick overview and refer tothe expository texts [16] by Kraft and [17] by Kraft and Riedtmann.

Let A = KQ/I be a finite-dimensional algebra and recall that A-modulescan be viewed as representations of Q. Moreover, each representationV = ((Vi)i∈Q0 , (Vα)α∈Q1) is isomorphic to a representation of the formM =

((Kdi)i∈Q0 , (Mα)α∈Q1

), where for each arrow α : i → j the matrix

Mα ∈ Kdj×di defines a linear map Kdi → Kdj .

For the sake of simplicity, we assume that the vertices of Q are 1, . . . , n.If we fix a dimension vector d ∈ Nn, then each A-module with dimensionvector d is isomorphic to a point in the space

rep(Q, d) =∏

(i→j)∈Q1

Kdj×di . (9.8)

If I 6= 0, then not every choice of matrices (Mα)α∈Q1 ∈ rep(Q, d) defines anA-module, but only those which satisfy the ideal I, see Section 3.8.

Example 9.18. Let Q =−→A 3 and I = 〈α2α1〉. Furthermore, let d ∈ N3 be

the dimension vector d1 = d2 = d3 = 2. Let M ∈ rep(Q, d) be any pair of

187


matrices R = Mα1 , S = Mα2 ∈ K2×2. Then (R,S) defines an A-module ifand only if SR = 0, equivalently if and only if the following four equationsare satisfied.

S11R11 + S12R21 = 0, S11R12 + S12R22 = 0,

S21R11 + S22R21 = 0, S21R12 + S22R22 = 0.

Note that each of these equation is polynomial. ♦

In general we denote by repI(Q, d) the subset of rep(Q, d) given by all fami-lies of matrices M = (Mα)α∈Q1 which satisfy the ideal I. The set repI(Q, d)is an algebraic variety: it is the set of all solutions of polynomial equationsin some affine space Kt. This is why repI(Q, d) is called module variety.

Two points M,N ∈ rep(Q, d) define isomorphic representations if and onlyif there exist linear invertible maps gi : K

di → Kdi , one for each vertex i,such that for any arrow α : i→ j in Q we have Nα gi = gj Mα. We canstate this differently: We define the group

GL(Q, d) =∏

i∈Q0

GL(di) (9.9)

and the group action of GL(Q, d) on rep(Q, d) by

g ·M = (gjMαg−1i )(α : i→j)∈Q1

. (9.10)

The orbit of M ∈ repI(Q, d) under the action of the group GL(Q, d) isdenoted by O(M). Then M and N are isomorphic if and only if they lie inthe same orbit under the action of GL(Q, d).

Example 9.19. Let A = KQ, where Q is the three-Kronecker quiver andd ∈ N2 given by d1 = d2 = 1. Then rep(Q, d) ' K3. Two points M,N ∈rep(Q, d) define isomorphic representations if and only if there exists someλ ∈ K \ 0 such that M = λN . Hence the orbits in rep(Q, d)/ ∼ underthe GL(Q, d)-action are given by P2(K) ∪ 0, that is, the projective spaceof dimension 2, where each of its point represents an isomorphism class ofindecomposable, together with a single point 0 representing the semisimpleS1 ⊕ S2. ♦

We now consider the dimensions of the space rep(Q, d) and of the groupGL(Q, d) presented in the previous section.

The dimension of rep(Q, d) is calculated easily:

dim rep(Q, d) =∑

(i→j)∈Q1

didj .

188


On the other hand GL(Q, d) is not a vector space, but an open set in G =∏i∈Q0

Kdi×di defined by a single polynomial inequality

(det g1)(det g2) · · · (det gn) 6= 0.

and therefore GL(Q, d) has the same dimension than G, that is,

dim GL(Q, d) =∑

i∈Q0

d2i .

The next result shows that these are exactly the ingredients of the quadraticform qKQ.

Proposition 9.20. Let Q be a finite quiver without cycle and A = KQ.Then for all dimension vectors d ∈ NQ0 we have

qA(d) = dim GL(Q, d)− dim rep(Q, d) (9.11)

Proof. The geometric form qA was defined in (8.6). Since A is hereditaryExt2

A(Si, Sj) = 0 and therefore

qA(d) =∑

i∈Q0

d2i −

∑(i→j)∈Q1

didj .

and hence the result follows.

Equation (9.11) is the first crucial equation, showing that the quadraticform expresses some geometric information. This also justifies the namegeometric form for the Tits form. The second crucial equation is stated inProposition 9.21, which we will not prove. It concerns for a fixed represen-tation M ∈ rep(Q, d) the stabilizer

GL(Q, d)M = g ∈ GL(Q, d) | g ·M = M,

which is a subgroup of GL(Q, d).

Proposition 9.21. Let Q be a finite quiver without cycle, d ∈ NQ0 andM ∈ rep(Q, d). Then the following equation holds

dimO(M) = dim GL(Q, d)− dim GL(Q, d)M . (9.12)

Note that the equality 9.12 is intuitively clear: The acting group changesM in some directions, whereas in others M remains unchanged, hence thedimension of the orbit is the dimension of the group minus the number ofdirections in which M remains fixed. We notice that GL(Q, d)M alwayscontains the one-dimensional subgroup consisting of the scalar multiples ofthe identity. Hence dim GL(Q, d)M ≥ 1.

189


Remark 9.22. We can use these dimension formulas to give an alterna-tive argument for Gabriel’s Thereom 9.5 as exposed in Section 9.2 which isalso known as the Tits argument. It shows that if A = KQ is of finiterepresentation type, then qA is positive definite.

Namely, assume that qA is not positive definite. Then there exists a non-zerovector d such that qA(d) ≤ 0. Write d = d+−d− with d±i ≥ 0 and d+

i d−i = 0

for any i. Then 0 ≥ qA(d) = qA(d+) + qA(d−)−∑qijd

+i d−j , where the sum

runs over all i, j such that d+i > 0 and d−j > 0. Since qij ≤ 0 we conclude

that qA(d) +∑qijd

+i d−j ≤ 0 and consequently qA(d+) ≤ 0 with d+ 6= 0 or

qA(d−) ≤ 0 with d− 6= 0. In any case, there exists a positive vector d suchthat qA(d) ≤ 0.

Hence by (9.11) and (9.12) we have

dim rep(Q, d) ≥ dim GL(Q, d)

= dimO(M) + dim GL(Q, d)M

> dimO(M)

and hence there can be no orbit in rep(Q, d) with the same dimension thanrep(Q, d) itself and consequently there must exist infinitely many orbits.This shows that A is not of finite representation type.

Note that in this argument it does not matter if these orbits considered inthe end correspond to indecomposable modules or not, since if A would beof finite representation type there were only finitely many orbits for anydimension vector. Unfortunately, this argument does not yield part (c) ofKac’s Theorem 9.10 as we cannot deduce from the existence of infinitelymany modules with dimension vector d that there are infinitely many inde-composables among them. ♦

Although it is considered “hopeless” to classify all indecomposables of a wildalgebra A it still makes sense to try to find some geometric “space” whichclassifies all A-modules of a fixed dimension vector (or a fixed dimension).This conducts to the notion of moduli, as defined by King in [15], seealso the overview [20] of Reineke. The notion of moduli attempts to viewrep(Q, d)/G(Q, d) as a geometrical quotient.

Exercises

9.6.1 Let A be the algebra considered in Example 9.18 and d ∈ N3 be given byd1 = d2 = d3 = 1. Calculate the orbits under the GL(Q, d)-action.

190


9.6.2 Let A = KQ/I, where Q =−→An and I is the ideal generated by all possible

compositions of two consecutive arrows. Let d ∈ Nn be given by di = m for all i.Calculate the dimension of rep(Q, d) as vector space. Furthermore determine howmany polynomial equations are needed to get repI(Q, d).

191


192

BIBLIOGRAPHY

Bibliography

[1] M. Auslander: Representation theory of Artin algebras II. Comm. Algebra 1(1974), 269–310.

[2] M. Auslander, I. Reiten, S. Smalø: Representation theory of Artin algebras.Corrected reprint of the 1995 original. Cambridge Studies in Advanced Math-ematics, 36. Cambridge University Press, Cambridge (1997), xiv+425pp. I

[3] R. Bautista, On algebras of strongly unbounded representation type. Com-ment. Math. Helv. 60 (1985), no. 3, 392–399.

[4] I. N. Bernstein, I. M. Gel’fand and V. A. Ponomarev: Coxeter functors andGabriel’s theorem. Russ. Math. Surv. 28 no. 2 (1973), 17–32.

[5] K. Bongartz, P. Gabriel, Covering spaces in representation-theory. Invent.Math. 65 (1981/82), no. 3, 331–378.

[6] W. Crawley-Boevey: Matrix problems and Drozd’s theorem. Topics in algebra,Part 1 (Warsaw, 1988), 199–222, Banach Center Publ., 26, Part 1, PWN,Warsaw, 1990.

[7] V. Dlab and C. M. Ringel: On algebras of finite representation type. J. Alge-bra 33 (1975), 306–394.

[8] V. Dlab and C. M. Ringel: Indecomposable representations of graphs andalgebras. Mem. Amer. Math. Soc. 173 (1976).

[9] J. Drozd: Tame and wild matrix problems. Representation theory II, 1980 -Springer.

[10] P. Gabriel: Unzerlegbare Darstellungen. I. (German) Manuscripta Math. 6(1972), 71–103; correction, ibid. 6 (1972), 309.

[11] P. Gabriel, L. A. Nazarova, A. V. Roiter, V. V. Sergeichuk and D. Vossieck:Tame and wild subspace problems. Ukranian Math. J. 45, Number 3, 335-372.

[12] J. E. Humphreys: Reflection groups and Coxeter groups. Cambridge Univer-sisty Press 29 (1990).

[13] V. Kac: Infinite root systems, representations of graphs and invariant theoryI. Invent. Math 56 (1980), 57–92.

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[14] V. Kac: Some remarks on representations of quivers and infinite root systems.Carleton Math. Lect. Notes 25 (1980), 1301–1317.

[15] A.D. King, Moduli of representations of finite-dimensional algebras. Quart.J. Math. Oxford Ser. (2) 45 (1994), no. 180, 515–530.

[16] H. Kraft: Geometric methods in representation theory. Lecture Notes inMathematics, 1982, Volume 944/1982, 180-258.

[17] H. Kraft, Ch.Riedtmann: Geometry of representations of quivers. In: Repre-sentations of algebras (ed. P. Webb) London Math. Soc. Lec. Note Series 116(1986), 109–145.

[18] L. A. Nazarova and A. V. Roiter: Representations of partially ordered sets.Zap. Nauchn. Sem. LOMI 28 (1972), 5-31. English translation: J. SovietMath. 23 (1975), 585-606.

[19] L. A. Nazarova and A. V. Roiter: Categorical matrix problems and the Brauer-Thrall conjecture. Preprint Inst. Math. Acad. Sci., Kiev 1973. German trans-lation: Mitt. Math. Sem. Giessen 115 (1975).

[20] M. Reineke: Moduli of representations of quivers. In: Trends in Representa-tion Theory of Algberas and Related Topics, EMS Publishing House 2008.Preprint arXiv:0802.2147v1.

[21] C.M. Ringel: On algorithms for solving vector space problems. I. Report on theBrauer-Thrall conjectures: Roiter’s theorem and the theorem of Nazarova andRoiter. Representation theory, I (Proc. Workshop, Carleton Univ., Ottawa,Ont., 1979), pp. 104–136, Lecture Notes in Math., 831, Springer, Berlin, 1980.

[22] C. M. Ringel: Tame algebras and integral quadratic forms. Lecture Notes inMath. 1099, Springer-Verlag (1984), xiii+376pp.

[23] A.V. Roiter: Unboundedness of the dimension of the indecomposable represen-tations of an algebra which has infinitely many indecomposable representa-tions. Izv. Akad. Nauk SSSR. Ser. Mat. 32 (1968), 1275-1282.

194

INDEX OF SYMBOLS

Index of Symbols

X←, X a vertex, 133X→, X a vertex, 133〈X,Y 〉, X,Y modules, 152〈x, y〉, x, y classes in K(A), 158[M ], M a module, 150

(x y)q, x, y dimension vectors, 164

(i i), i a vertex, 26p ∼ q, p, q unit forms, 160Mn, M a module, n an integer, 41Aop, A an algebra, 63C op, C a category, 63C ×D , C ,D two categories, 94M>, M a matrix, 10V n, V a representation, n an inte-

ger, 210, (zero module), 38, 640, (zero morphism), 640, (zero matrix), 21r, (identity matrix), 2ΓA, A an algebra, 107∆(q), q a unit form, 159δX,Y , X,Y indecomposable mod-

ules, 133νA, A an algebra, 78, 121τA, A an algebra, 119τ−A , A an algebra, 119ΦA, A an algebra, 155χA, A an algebra, 158Ψ(q), q a unit form, 165An, n an integer, 159An, n an integer, 166AQ, Q a quiver, 40−→An, n an integer, 31add D , D a subcategory of C , 107

CA, A an algebra, 153Coim f , f a morphism, 67Coker f , f a morphism, 66Dn, n an integer, 159Dn, n an integer, 166DV , V a vector space, 63dimM , M a representation, 19E (Z,X), Z and X two modules, 86En, n an integer, 159En, n an integer, 166ei, i a vertex, 26, 151EndA, A an algebra, 39Ext1

A, A an algebra, 86ExtnA, A an algebra, n an integer,

100gldimA, A an algebra, 100HomA, A an algebra, 38HomA, A an algebra, 124HomA, A an algebra, 124HomC , C a category, 27HomQ, Q a quiver, 24Ix, x a vertex, 76idimM , M a module, 100Im f , f a morphism, 67indA, A an algebra, 107Jn, n an integer, 110J(n, λ), n ≥ 0 an integer, 11J(m, 0), m ≥ 0 an integer, 10K(A), A an algebra, 150Ker f , f a morphism, 66len(w), w a path in a quiver, 26MQ, Q a quiver, 19MQ,d, Q a quiver, d ∈ NQ0 , 19modA, A an algebra, 38

195

INDEX OF SYMBOLS

mod C , C a K-category, 27ModA, A an algebra, 64modA, A an algebra, 124modA, A an algebra, 124Qop, Q a quiver, 64P (q), q a unit form, 163Px, x a vertex, 74pdimM , M a module, 100projA, A an algebra, 124QC , C a spectroid, 53Qn, n an integer, 110q(∆), ∆ a diagram, 159R(q), q a unit form, 163Rn,λ, n an integer, λ ∈ K, 110rad C , C a K-category, 52rad∞A , A an algebra, 108repQ, Q a quiver, 24rep(Q, d), Q a quiver, d a dimension

vector, 187repI(Q, d), Q a quiver, d a dimen-

sion vector, I ⊂ KQ anideal, 188

Sx, x a vertex, 82socM , M a module, 83supp v, v a dimension vector, 179

196

INDEX OF NOTIONS

Index of Notions

abeliancategory, 68, 68, 88, 127

with enough projectives and injec-tives, 80, 85

additivecategory, 67, 68, 78closure, 107

adjacency matrix, 28admissible ideal, 55, 55, 56–59, 146, 180, 183algbera

its quiver, 76algebra, 13, 37, 37, 38, 39, 59, 61–64, 71–74,

76, 78–80, 82, 85, 100, 101, 107,109, 117, 125, 127, 129, 134, 136,138, 139, 144, 151–154, 156, 158,162, 167, 170

basic, 47, 48, 53, 55, 58, 144connected, 109, 138, 144, 145, 173, 174hereditary, 144, 144, 146, 153, 156,

172, 174, 175, 189homomorphism, 38, 38, 39, 43, 116isomorphism, 38, 39–43, 46, 56, 72,

108, 116its center, 37, 37, 63its dimension, 38, 38, 39, 40, 44its quiver, 47, 48, 53, 56, 58, 59, 64, 78,

81, 83, 101, 109, 121, 133, 135,136, 144, 145, 151, 153, 155, 169,185

its radical, 49, 49, 51, 52, 58, 138Kronecker, 40, 42, 95, 109, 114, 146,

157, 167, 184local, 45, 46, 47, 49, 51–53, 70, 74, 126,

129of endomorphisms, 39, 72, 138of paths, 40, 40, 41–43, 55, 56, 108,

116, 135, 144opposite, 63polynomial, 41, 185quotient, 42, 42, 49, 56, 57, 108three subspace, 138two subspace, 41, 109, 120, 138

algebraic variety, 188almost split sequence, 127

AR, see Auslander-Reitenarrow, 18, 18, 19, 20, 26, 28–31, 34, 41–43,

53–55, 63, 64, 75, 78, 81, 83, 87,101, 108, 110, 112, 114, 115, 121,122, 133, 138, 144–146, 153, 159,163, 175, 183, 184, 187, 188, 191

ending in, 18, 133, 175starting in, 18, 133, 175

Auslander’s Theorem, 138Auslander-Reiten

orbit, 139, 139, 140, 143–146quiver, 107, 107, 108–110, 120, 124,

127, 129, 133, 137–140, 143, 157Kronecker, 109, 112, 114linearly oriented, 120

sequence, 121, 124, 125–127, 129, 131,132, 135–137, 146, 157

theorem, 125, 127, 129translate, 117, 118, 119, 120, 124, 139,

157

Baer sum, 90, 90, 92basic, 47, 48, 53, 55, 58, 144basis

of a free module, 149of a representation, 20, 20, 21, 29, 75

BGP-reflection functor, 176bifunctor, 94bilinear form, 152, 153, 158, 164, 176–178

homological, 153, 158bimodule, 39, 54, 185bounded

quiver, 56, 59representation type, 169, 169, 171

bounded quiver, 74Brauer-Thrall conjecture, 169, 171, 172,

172

Cartan matrix, 153, 153, 154, 155, 163category, 22, 22, 23–29, 31, 33, 51, 58, 59,

62, 64–67, 71, 72, 78K-category, 27, 27, 28, 42–44, 52, 53,

58, 73abelian, 68, 68, 88, 127

197

INDEX OF NOTIONS

with enough projectives and injec-tives, 80, 85

additive, 67, 68, 78composition within, 22, 23, 24, 27, 29,

31, 42–44, 57, 63, 67, 94concrete, 62, 63dual, 25equivalent, 25, 29, 29, 31, 45, 48, 55,

59, 77, 109, 179, 180, 183isomorphic, 25, 25, 28, 29, 31, 40, 41,

43, 55, 57, 63, 64, 71, 72, 78, 117its radical, 52, 58morphism, 22, 23–29, 31, 42–44, 51–

53, 55, 65, 94of K-linear functors, 27of modules, 38, 40, 42, 48, 61, 63–68,

71, 85, 109, 118, 121, 124stable, 124

of paths, 26, 27, 27, 40, 56of representations, 24, 29opposite, 63, 63, 64product, 94quotient, 42, 43, 124, 127skeleton, 71, 71, 72small, 71, 71spectroid, 51, 51, 52, 53, 55, 59, 76,

78, 107subcategory, 57, 57, 68, 71, 77, 107,

180center, 37, 37, 63class of a module, 150, 151–154, 156closed under

direct sums, 107isomorphisms, 107

coimage, 67, 68cokernel, 64, 65, 65, 66–68, 88, 90, 134–136,

145, 175complete set of idempotents, 44, 44, 45, 47,

48, 51, 53, 55, 64, 73, 74, 76, 78,82

component, 108, 108, 114, 138, 143, 145,166, 167

finite, 108, 138, 144infinite, 108, 143preinjective, 114, 143, 144–146, 157preprojective, 114, 143, 144–146, 157,

167, 174, 175regular, 115, 143

compositionfactors, 82, 83in a category, 22, 23, 24, 27, 29, 31,

42–44, 57, 63, 67, 94of functors, 25, 27of paths, 27, 27, 40, 54, 59series, 82

concrete, 62, 63

connectedalgebra, 109, 138, 144, 145, 173, 174

connecting map, 103, 173contravariant, 24, 25, 26, 59, 94, 127corank, 160, 162coupled

4-block, 8, 8, 9, 21transformations, 7, 8, 9, 11–13

covariant, 24, 24, 25–28, 56, 59, 94cover, 79, 79, 80–82, 97, 118, 119Coxeter

matrix, 155reflection functor, 176, 179reflections, 178, 178, 179transformation, 155, 155, 157

cycle, 26, 28, 143–146, 153, 173–176, 179,180, 189

decomposition into indecomposables, 5, 6,11, 12, 32, 69, 70, 70, 73, 82, 121,126, 132, 174

degreeof a vertex, 167, 167

diagramDynkin, 159, 159, 161–167, 173, 178,

179exceptional, 159extended, 162, 162

of roots, 165, 165, 167dimension

finitistic, 100global, 100, 152, 154, 155, 157, 158injective, 100, 100of a module, 38, 42, 69, 80–82, 86, 90,

95, 96, 105, 141, 151, 155, 169,170, 172, 184, 187, 190

of an algebra, 38, 38, 39, 40, 44projective, 100, 100, 101vector, 19, 19, 20, 25, 96, 136–138,

140–143, 149, 156, 157, 163, 167,169, 174–178, 180, 185–187, 189,190

direct sumin a category, 28, 28, 73, 107modules, 41of elements in a matrix problem, 28of elements of a matrix problem, 7, 8,

11, 12of elements of a matrix problems, 5of modules, 67, 69, 70, 73, 74, 76, 78,

79, 83, 121, 123, 125, 133, 135,138

of representations, 21, 28, 32, 33, 176,177

of short exact sequences, 90representations, 27

198

INDEX OF NOTIONS

dotted edge, 159, 159, 165, 165, 167dual

category, 25matrix problem, 6, 17notions, 62, 63, 65, 66, 68, 72, 80, 90,

118, 130space, 63

dualization, 25, 63, 66, 69, 80, 89, 90, 92,94, 100, 119, 122, 126, 127, 131,132, 136, 143–145

Dynkindiagram, 159, 159, 161–167, 173, 178,

179exceptional, 159extended, 162, 162

quiver, 179, 179

edge, 159, 165dotted, 159, 159, 165, 165, 167full, 159, 165, 167

embedding, 182, 182, 183end terms, 85, 124, 136, 137endomorphism

algebra, 39, 51, 72, 138of a module, 39, 39, 45, 52, 71, 79, 126ring, 39

epi, see epimorphismepimorphism, 62, 62, 63, 66, 67, 72, 73, 78–

80, 82split, 78, 79, 109, 132

equivalence, 29Morita, 47, 48, 48, 49, 53, 55, 58, 180of categories, 25, 29, 29, 31, 45, 48, 55,

59, 77, 109, 179, 180, 183of elements in a matrix problem, 1, 2–

7, 12, 18, 20, 24of functors, 29of short exact sequences, 86, 88, 89,

92, 95, 98of unit forms, 160, 167

Euler form, 158, 158, 162, 173exact

functor, 127, 127sequence, 96, 98, 103, 104, 118, 118,

120–122, 124–126, 152, 155, 156,172, 173

long, 101, 104, 104, 105, 105, 152short, 85, 85, 86–92, 95, 98, 101,

103, 105, 124, 126, 127, 136, 150–152, 172, 174, 176

exceptional, 159extended Dynkin diagram, 162, 162extension group, 86, 95, 96, 100, 101, 103,

153higher, 100

factor over, 65, 68, 73, 79, 94, 96, 98, 100,102, 104, 109, 115, 116, 122, 124,126, 127, 132, 138, 146, 179

factor through, 65, 130–132factorization

of a morphism, 68, 73, 79, 94, 96, 98,100, 102, 104, 109, 115, 116, 122,124, 126, 127, 130–132, 138, 146,179

faithful, 182, 182fatorization

of a morphism, 65finite

component, 108, 138, 144quiver, 18, 19–21, 23, 25, 27–29, 40,

55, 56, 62, 146, 175, 176, 179,180, 189

representation type, 14, 32, 55, 107,108, 171–174, 186, 190

spectroid, 51, 51, 52, 53, 55, 59, 76finitely generated free

module, 149, 185finitistic dimension, 100form

bilinear, 152, 153, 158, 164, 176–178Euler, 158, 158, 162, 173geometric, 158, 189normal, 1, 3, 4, 6, 7, 11quadratic, 158, 160, 161, 163, 176, 189reduced row-echelon, 3Ringel, 158Tits, 158, 172, 177, 189unit, 158, 159, 160, 162–168, 178, 179

four subspace quiver, 143free

finitely generated, 149, 185module, 149, 149, 150, 151, 154

its basis, 149full

edge, 159, 165, 167functor, 182, 182, 183, 184subcategory, 57, 57, 59, 68, 71, 71, 77,

107, 180subquiver, 144, 179

functor, 22, 24, 24, 25–27, 29, 30, 48, 55,57, 59, 74, 94, 117, 123, 127, 172,175, 176, 181–184, 187

bifunctor, 94composition, 25, 27contravariant, 24, 25, 26, 59, 94, 127covariant, 24, 24, 25–28, 56, 59, 94exact, 127, 127faithful, 182, 182full, 182, 182, 183, 184identity, 25isomorphism, 25, 26, 29, 31, 49

199

INDEX OF NOTIONS

linear, 27, 28, 45, 56, 59, 94, 94morphism, 25, 26–28, 30, 45, 59of Nakayama, 78, 119–121, 121, 122quasi-inverse, 29, 30reflection, 175, 176, 176, 178, 179

fundamental region, 179, 180

Gabriel’s Theorem, 55, 172, 173, 173, 190generating set, 50, 50geometric form, 158, 189global dimension, 100, 152, 154, 155, 157,

158graph

of a unit form, 159, 161–163underlying a quiver, 159, 159, 167

Grothendieck group, 149, 150, 152

hereditary, 144, 144, 146, 153, 156, 172, 174,175, 189

homologicalbilinear form, 152, 153, 158quadratic form, 158, 158

homomorphismimage, 62, 65, 81, 87of algebras, 38, 38, 39, 43, 116of modules, 38, 38, 41, 45, 46, 48, 50,

59, 61, 70, 72, 86–91, 93, 94, 97–100, 103, 117, 121, 122, 124, 129–132, 134, 136, 138, 144–146, 149–151

radical, 72, 108, 112, 124, 126, 129–132, 136, 138, 140

hullinjective, 78, 80, 80, 125

idealadmissible, 55, 55, 56–59, 146, 180,

183nilpotent, 51, 51, 52, 52, 55, 108, 138of a category, 42, 42, 43, 52, 53, 55, 56,

124, 179of an algebra, 42, 42, 43, 47, 49–52,

55–57, 97, 126, 139, 143, 155, 163,191

power, 51, 52, 54satisfied by, 56, 57–59, 180, 187, 188zero, 52

idempotent, 43, 44, 44, 46, 47, 59, 177, 182complete set, 44, 44, 45, 47, 48, 51, 53,

55, 64, 73, 74, 76, 78, 82orthogonal, 44, 44, 45, 47, 48, 51, 53,

55, 64, 73, 74, 76, 78, 82, 183primitive, 44, 44, 45–48, 51, 53, 55, 64,

73, 74, 76, 78, 82identity

functor, 25

matrix, 2, 2, 7, 24, 29, 30morphism, 22, 23, 24, 26, 29, 31, 94,

126of functors, 25

path, 26image, 68

of a homomorphism, 62, 65, 81, 87of a morphism, 67, 67, 68

imaginary Schur root, 180, 180indecomposable

element of a matrix problem, 5, 5, 6,7, 12–14, 19, 21

module, 41, 41, 42, 46, 47, 58, 69–74,76, 78, 81, 82, 100, 107–109, 119–122, 124–126, 129–140, 142–146,153–157, 163, 167, 169, 172, 174,175, 180, 181, 184, 186–188, 190

of the Kronecker algebra, 109, 110,110, 111, 112, 117

of the Kronecker problem, 7–11, 11,12, 19, 20, 22

of the two subspace problem, 21parameter family of, 13representation, 20, 21, 21, 22, 34, 57,

58, 74, 81, 100, 109, 172, 176,178–181

of the linearly oriented quiver, 31–34

indefinite, 160index of inertia, 160indivisible, 181infinite

component, 108, 143radical, 107, 108representation type, 14, 108, 172

initial object, 62, 63, 64injective

dimension, 100, 100enough, 80, 85hull, 78, 80, 80, 125module, 76object, 72, 72, 76–78, 96, 119, 120,

124, 125, 131, 134–136, 139–145presentation, 118, 144

minimal, 119resolution, 96, 100, 125

irreduciblemodule, 81morphism, 34, 34, 35, 53, 57, 71, 108,

112, 115, 145, 146isomorphism

in a category, 24, 26, 28, 47–49, 51, 52,66, 68, 69, 71, 74, 107

of algebras, 38, 39–43, 46, 56, 72, 108,116

200

INDEX OF NOTIONS

of categories, 25, 25, 28, 29, 31, 40, 41,43, 55, 57, 63, 64, 71, 72, 78, 117

of functors, 25, 26, 29, 31, 49of modules, 42, 46, 58, 61, 67, 68, 70,

71, 73, 74, 76, 78–83, 86, 88, 95,99, 107, 108, 118–121, 129–133,136, 137, 140, 150, 151, 169, 183,186, 187

of quivers, 58, 145of representations, 20, 20, 21, 26, 30–

34, 57, 58, 75, 95, 110, 112, 176–178, 180–182, 187, 188

of unit forms, 160

Jacobson radical, 49, 51, 52, 57, 58Jordan-Holder filtration, 82, 151

kernel, 64, 65, 65, 66–69, 90, 97, 102, 119,122, 123, 132, 142

knitting, 133, 134, 136–143Kronecker

algebra, 40, 42, 95, 109, 114, 146, 157,167, 184

its indecomposables, 109, 110, 110,111, 112, 117

Auslander-Reiten quiver, 109, 114matrix problem, 7, 7, 10–12, 14, 17,

19, 20, 109its indecomposables, 7–11, 11, 12,

19, 20, 22quiver, 18, 20, 22, 28, 40, 109, 116, 184

Krull-Remak-Schmidt’s Theorem, 70

Lemmaof Nakayama, 50, 51, 82of Schur, 81, 82, 153of Yoneda, 100, 144

lengthof a path, 26, 26, 28, 53of a projective resolution, 100, 154

linearfunctor, 45, 56, 59, 94, 94

linearly oriented quiver, 31, 42, 43, 57, 80,108, 159

indecomposable representations, 31, 32local, 45, 46, 47, 49, 51–53, 70, 74, 126

algebra, 129long exact sequence, 101, 104, 104, 105,

105, 152loop, 18, 26, 41–43, 71, 159, 180, 181

matrixadjacency, 28Cartan, 153, 153, 154, 155, 163Coxeter, 155identity, 2, 2, 7, 24, 29, 30

zero, 2, 6, 113matrix problem, 1, 14, 15, 21, 22, 29, 37, 139

conjugation, 2, 12coupled 4-block, 8, 8, 9, 21direct sum, 5, 7, 8, 11, 12, 28dual, 6, 17equivalent elements, 1, 2–7, 12, 18, 20,

24Kronecker, 7, 7, 10–12, 14, 17, 19, 20,

109three Kronecker, 12, 13, 14, 17three subspace, 7, 14, 19two subspace, 3, 4–7, 12, 14, 17, 21

middle term, 85, 86, 88, 95, 96, 136, 146minimal

injective presentation, 119projective presentation, 118, 118, 119,

120, 125, 142module, 13, 37, 38, 38, 39–41, 45–47, 50,

51, 56, 58, 59, 61–64, 68–74, 78–80, 83, 85, 86, 91, 93, 94, 96, 100,101, 103, 105, 114, 117–119, 121,122, 124, 125, 136, 138, 140, 142–144, 149, 151, 152, 154, 155, 172,174, 184–188, 190

category, 38, 40, 42, 48, 61, 63–68, 71,85, 109, 118, 121, 124

stable, 124direct sum, 41, 67, 69, 70, 73, 74, 76,

78, 79, 83, 121, 123, 125, 133,135, 138

finitely generated free, 149, 185free, 149, 149, 150, 151, 154generated by, 50, 50homomorphism, 38, 38, 41, 45, 46, 48,

50, 59, 61, 70, 72, 86–91, 93, 94,97–100, 103, 117, 121, 122, 124,129–132, 134, 136, 138, 144–146,149–151

indecomposable, 41, 41, 42, 46, 47, 58,69–74, 76, 78, 81, 82, 100, 107–109, 119–122, 124–126, 129–140,142–146, 153–157, 163, 167, 169,172, 174, 175, 180, 181, 184, 186–188, 190

injective, 76, 78, 96, 119, 120, 124, 125irreducible, 81isomorphism, 42, 46, 58, 61, 67, 68, 70,

71, 73, 74, 76, 78–83, 86, 88, 95,99, 108, 118–121, 129–133, 136,137, 140, 150, 151, 169, 183, 186,187

its class, 150, 151–154, 156its dimension, 38, 42, 69, 80–82, 86, 90,

95, 96, 105, 141, 151, 155, 169,170, 172, 184, 187, 190

201

INDEX OF NOTIONS

its endomorphisms, 39, 39, 45, 52, 71,79, 126

its radical, 50, 51, 127, 135–137, 139,141

object, 78preinjective, 143, 146, 156, 163preprojective, 143, 156projective, 72, 72, 73–76, 78–82, 96,

101–103, 118–127, 131–133, 135–141, 143–146, 153, 154, 156, 163

quotient, 38, 62, 82, 144regular, 143, 146semisimple, 83, 83, 188simple, 81, 81, 82, 105, 137, 144, 151,

153, 158submodule, 38, 38, 46, 50, 51, 62, 65,

81–83, 140, 144, 151variety, 188zero, 38, 41, 64, 81, 82, 119, 120, 136,

182, 183moduli, 190mono, see monomorphismmonomorphism, 62, 62, 63, 66–68

split, 109, 119Morita equivalence, 47, 48, 48, 49, 53, 55,

58, 180morphism

coimage, 67, 68cokernel, 65, 65, 66–68, 88, 90, 134–

136, 145, 175epi, 62, 62, 63, 66, 67, 72, 73, 78–80,

82factoring over, 68, 73, 79, 94, 96, 98,

100, 102, 104, 109, 115, 116, 122,124, 126, 127, 132, 138, 146, 179

factoring through, 65, 130–132identity, 22, 23, 24, 26, 29, 31, 94, 126

of functors, 25image, 67, 67, 68in a category, 22, 23–29, 31, 42–44, 51–

53, 55, 65, 94irreducible, 34, 34, 35, 53, 57, 71, 108,

112, 115, 145, 146kernel, 65, 65, 66–69, 90, 97, 102, 119,

122, 123, 132, 142mono, 62, 62, 63, 66–68of categories, 59of functors, 25, 26–28, 30, 45, 59of representations, 20, 20, 24–26, 28–

30, 33–35, 41, 68, 176, 179, 182–184

radical, 52, 52, 53zero, 33, 33, 34, 39, 62, 64, 81, 110–

112, 115, 117, 136, 138, 140, 146,157, 173

Nakayama

functor, 78, 119–121, 121, 122Lemma, 50, 51, 82

natural transformation, 25negative

index of inertia, 160root, 163, 164

nilpotentelement, 28, 45, 46, 47, 49, 52, 74, 129ideal, 51, 51, 52, 52, 55, 108, 138

normal form, 1, 3, 4, 6, 7, 11

objectdirect sum, 28, 28, 73, 107initial, 62, 63, 64injective, 72, 72, 76–78, 131, 134–136,

139–145isomorphic, 24, 26, 28, 47–49, 51, 52,

66, 68, 69, 71, 74, 107projective, 72, 72, 73, 76–78terminal, 62, 63, 64zero, 64, 64, 65–67, 127

one-parameter family, 12, 184, 185, 185, 186opposite

algebra, 63category, 63, 63, 64quiver, 64ring, 39

orbitAuslander-Reiten, 139, 139, 140, 143–

146in a module variety, 188, 188, 190

orthogonal idempotents, 44, 44, 45, 47, 48,51, 53, 55, 64, 73, 74, 76, 78, 82,183

path, 26, 26, 27, 28, 40, 41, 53, 56, 57, 116,145, 146

algebra, 40, 40, 41–43, 55, 56, 108, 116,135, 144

category, 26, 27, 27, 40, 56composition, 27, 27, 40, 59identity, 26its length, 26, 26, 28, 53trivial, 26

positivedefinite, 160, 160, 161–163, 166, 167,

173, 174, 178, 190index of inertia, 160root, 163, 163, 164–166, 168, 173, 174,

178, 180semidefinite, 160, 160, 162, 162

power of an ideal, 51, 52, 54preinjective

component, 114, 143, 144–146, 157module, 143, 146, 156, 163

preprojective

202

INDEX OF NOTIONS

component, 114, 143, 144–146, 157,167, 174, 175

module, 143, 156presentation

injective, 118, 144minimal, 119

projective, 118, 118, 119, 143, 144, 155minimal, 118, 118, 119, 120, 125,

142preserve indecomposabilitiy, 181, 181, 182–

184, 187primitive idempotent, 44, 44, 45–48, 51, 53,

55, 64, 73, 74, 76, 78, 82product category, 94projective

cover, 78, 79, 79, 80–82, 97, 118, 119dimension, 100, 100, 101module, 81object, 72, 72, 73–80, 82, 96, 101–103,

118–127, 131–133, 135–141, 143–146, 153, 154, 156, 163

presentation, 118, 118, 119, 143, 144,155

minimal, 118, 118, 119, 120, 125,142

resolution, 96, 97, 99–101, 154, 155pull-back, 68, 68, 69, 89, 90, 92, 93, 100, 126push-out, 68, 69, 87–91, 93

quadratic form, 158, 160, 161, 163, 176, 189Euler, 158, 158, 162, 173geometric, 158, 189homological, 158, 158Ringel, 158Tits, 158, 172, 177, 189

quasi-inverse functors, 29, 30quiver, 15, 17, 18, 18, 19, 21, 22, 24, 26,

28, 29, 31, 34, 37, 40–43, 47, 53,54, 56–59, 68, 71, 75, 78, 96, 108,109, 120, 138, 139, 141, 144–146,153, 167, 173, 175, 178–181, 183,185, 196

Auslander-Reiten, 107, 107, 108–110,112, 114, 120, 124, 127, 129, 133,137–140, 143, 157

bounded, 56, 59, 74Dynkin, 179, 179finite, 18, 19–21, 23, 25, 27–29, 40, 55,

56, 62, 146, 175, 176, 179, 180,189

four subspace, 143isomorphism, 58, 145Kronecker, 18, 20, 22, 28, 40, 109, 116,

184linearly oriented, 31, 31, 32, 42, 43, 57,

80, 108, 159

of a spectroid, 53, 55of an algebra, 47, 48, 53, 56, 58, 59,

64, 76, 78, 81, 83, 101, 109, 121,133, 135, 136, 144, 145, 151, 153,155, 169, 185

opposite, 64subquiver, 144, 146

full, 144, 179three Kronecker, 18, 22, 188three subspace, 19two subspace, 18, 41, 78, 81

quotientalgebra, 42, 42, 49, 56, 57, 108category, 42, 43, 124, 127module, 38, 62, 82, 144

radical, 49homomorphism, 72, 108, 112, 124, 126,

127, 129–132, 136, 138, 140infinite, 107, 108Jacobson, 49, 51, 52, 57, 58morphism, 52, 52, 53of a category, 52, 58of a module, 50, 51, 135–137, 139, 141of a spectroid, 52, 53–55, 108of an algebra, 49, 49, 51, 52, 58, 138

real Schur root, 180, 180reduced row-echelon form, 3reflect isomorphisms, 181, 181, 182–184, 187reflection

function, 178, 178, 179functor, 175, 176, 176, 178, 179

regularcomponent, 115, 143module, 143, 146

relation, 56, 56, 57, 74, 75, 141, 146, 150,155

zero, 56, 97representation

basis, 20, 20, 21, 29, 75direct sum, 21, 27, 28, 32, 33, 176, 177indecomposable, 20, 21, 21, 22, 32–34,

57, 58, 74, 81, 100, 109, 172, 176,178–181

isomorphism, 20, 20, 21, 26, 30–34,57, 58, 75, 95, 110, 112, 176–178,180–182, 187, 188

morphism, 20, 20, 24–26, 28–30, 33–35, 41, 68, 176, 179, 182–184

of a quiver, 15, 19, 20–22, 24–27, 29,31, 33, 34, 37, 40, 41, 56–59, 62,64, 68, 74, 75, 80, 81, 83, 95, 96,110, 120, 136, 151, 175–178, 180,181, 183–185, 187, 189

satisfying an ideal, 56, 57–59, 180, 187,188

203

INDEX OF NOTIONS

subrepresentation, 34zero, 19, 21, 32

representation type, 14bounded, 169, 169, 171finite, 14, 32, 55, 107, 108, 171–174,

186, 190infinite, 14, 108, 172strongly unbounded, 170, 172tame, 14, 186unbounded, 170, 172wild, 13, 14, 181, 190

resolutioninjective, 96, 100, 125projective, 96, 97, 99–101, 154, 155

restriction, 160, 161, 162retraction, 109, 124, 126Ringel form, 158root, 163, 163, 164–167, 172, 174, 178, 179

diagram, 165, 165, 167neagtive, 164negative, 163, 164positive, 163, 163, 164–166, 168, 173,

174, 178–180Schur, 180, 180

imaginary, 180, 180real, 180, 180

satisfy an ideal, 56, 57–59, 180, 187, 188Schur

Lemma, 81, 82, 153root, 180, 180

imaginary, 180, 180real, 180, 180

section, 109, 124, 127, 132, 140, 155semisimple, 83, 83, 188sequence

almost split, 127Auslander-Reiten, 121, 124, 125–127,

129, 131, 132, 135–137, 146, 157exact, 96, 98, 103, 104, 118, 118, 120–

122, 124–126, 152, 155, 156, 172,173

long exact, 101, 104, 104, 105, 105,152

short exact, 85, 85, 86–92, 95, 98, 101,103, 105, 124, 126, 127, 136, 150–152, 172, 174, 176

short exact sequence, 85, 85, 86–92, 95, 98,101, 103, 105, 124, 126, 127, 136,150–152, 172, 174, 176

direct sum, 90end terms, 85, 124, 126, 136, 137ending in, 85, 86, 88, 127, 135, 138,

146, 157equivalent, 86, 88, 89, 92, 95, 98middle term, 85, 86, 88, 95, 96, 136,

146

split, 92, 95, 124, 126, 127, 132, 150,172, 174

starting in, 85, 86, 137, 138short exact sequences

sum, 90, 92simple module, 81, 81, 82, 105, 137, 144,

151, 153, 158sink, 133, 135, 145, 175, 176, 178, 179

map, 129, 130, 130, 131–136, 138sink map, 131skeleton, 71, 71, 72small, 71, 71socle, 83source, 133, 145, 175–179

map, 129, 129, 130–136, 140, 145source map, 131spectroid, 51, 51, 52, 53, 55, 59, 74, 78, 107

finite, 51, 51, 52, 53, 55, 59, 76its quiver, 53its radical, 52, 53–55, 108

splitepi, 78, 79, 109, 132mono, 109, 119short exact sequence, 92, 95, 124, 126,

127, 132, 150, 172, 174stable module catgeory, 124starting vertex, 18, 18, 56strongly unbounded

representation type, 170, 172subcategory, 57, 57

full, 57, 57, 59, 68, 71, 71, 77, 107, 180submodule, 38, 38, 51, 62, 65, 81–83, 140,

144, 151subquiver, 144, 146

full, 144, 179subrepresentation, 34sum

Baer, 90, 90, 92of short exact sequences, 90, 90, 92

support, 179Sylvester’s index of inertia, 160

tame, 14, 186terminal object, 62, 63, 64terminating vertex, 18, 18, 56Theorem

of Auslander, 138of Auslander-Reiten, 125, 127, 129of Gabriel, 55, 172, 173, 190of Krull-Remak-Schmidt, 70

three Kroneckermatrix problem, 12, 13, 14, 17quiver, 18, 22, 188

three subspacealgebra, 138matrix problem, 7, 14, 19

204

INDEX OF NOTIONS

quiver, 19Tits

argument, 190form, 158, 172, 177, 189

transformationallowed, 3, 7, 8, 10coupled, 7, 8, 9, 11–13

translate of Auslander-Reiten, 117, 118,119, 120, 124, 139, 157

trivial path, 26two parameter family, 13two subspace

algebra, 41, 109, 120, 138matrix problem, 3, 4–7, 12, 14, 17, 21

its indecomposables, 21quiver, 18, 41, 78, 81

typerepresentation, 14

bounded, 169, 169, 171finite, 14, 32, 55, 107, 108, 171–174,

186, 190infinite, 14, 108, 172strongly bounded, 170strongly unbounded, 172tame, 14, 186unbounded, 170, 172wild, 13, 14, 181, 190

unboundedrepresentation type, 170, 172

underlying graph, 159, 159, 167, 173, 179unit form, 158, 159, 160, 162–168, 178, 179

equivalent, 160, 167isomorphic, 160its graph, 159, 161–163

universal propertyof free modules, 149

variety, 188vertex, 18, 18, 19, 20, 26–31, 41–43, 47, 48,

53, 54, 57, 62, 64, 68, 71, 74, 81,82, 95, 108, 114, 117, 120, 121,129, 133, 135, 140, 144–146, 151,159, 160, 162, 164, 165, 167, 175,177, 179, 181, 185, 187, 188

its degree, 167, 167starting, 18, 18, 56terminating, 18, 18, 56

Weyl group, 179, 179, 180wild, 13, 14, 181, 184, 190

Yoneda’s Lemma, 100, 144

zeroideal, 52

matrix, 2, 6, 113module, 38, 41, 64, 81, 82, 119, 120,

136, 182, 183morphism, 33, 33, 34, 39, 62, 64, 81,

110–112, 115, 117, 136, 138, 140,146, 157, 173

object, 64, 64, 65–67, 127relation, 56, 97representation, 19, 21, 32

205

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